Survey of Geometry. Supplementary Notes on Elementary Geometry. Paul Yiu. Department of Mathematics Florida Atlantic University.

Transcription

1 Survey of Geometry Supplementary Notes on Elementary Geometry Paul Yiu Department of Mathematics Florida tlantic University Summer 2007

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3 ontents 1 The Pythagorean theorem i 1.1 The hypotenuse of a right triangle i 1.2 The Pythagorean theorem iii 1.3 Integer right triangles iii 2 The equilateral triangle v 2.1 onstruction of equilateral triangle v 2.2 The medians of an equilateral triangle vi 2.3 The center of an equilateral triangle vii 2.4 The circumcircle and incircle of an equilateral triangle vii 2.5 Trigonometry viii 2.6 Some interesting examples on equilateral triangles x 3 The square xiii 3.1 onstruction of a square on a segment xiii 3.2 The diagonals of a square xiv 3.3 The center of the square xv 3.4 Some interesting examples on squares xv 4 Some basic principles xvii 4.1 ngle properties xvii Parallel lines xvii ngle sum of a triangle xviii ngle properties of a circle xviii 4.2 Tests of congruence of triangles xix onstruction of a triangle with three given elements... xix ongruence tests xix

4 iv ONTENTS 5 ircumcircle and incircle xxiii 5.1 ircumcircle xxiii The perpendicular bisector locus theorem xxiii onstruction of circumcircle xxiv ircumcircle of a right triangle xxiv 5.2 The incircle xxv The angle bisector locus theorem xxv onstruction of incircle xxv The incircle of a right triangle xxvi 5.3 Tangents of a circle xxvi Tangent at a point on the circle xxvi The tangents from a point to a circle xxvi 5.4 Some interesting examples on the circumcircle and incircle of a triangle xxvii 6 Uses of congruence tests xxix 6.1 Isosceles triangles xxix 6.2 hords of a circle xxx 6.3 Parallelograms xxxi Properties of a parallelogram xxxi Quadrilaterals which are parallelograms xxxii Special parallelograms xxxii 6.4 The midpoint theorem and its converse xxxiii The midpoint theorem xxxiii The converse of the midpoint theorem xxxiv Why are the three medians concurrent? xxxiv 6.5 Some interesting examples xxxv 7 The regular hexagon, octagon and dodecagon xxxvii 7.1 The regular hexagon xxxvii 7.2 The regular octagon xxxviii 7.3 The regular dodecagon xxxix 8 Similar triangles xli 8.1 Tests for similar triangles xli 8.2 The parallel intercepts theorem xliii 8.3 The angle bisector theorem xliv 8.4 Some interesting examples of similar triangles xlv

5 ONTENTS v 9 Tangency of circles xlix 9.1 External and internal tangency of two circles xlix 9.2 Three mutually tangent circles l 9.3 Some interesting examples of tangent circles l 10 The regular pentagon liii 10.1 The golden ratio liii 10.2 The diagonal-side ratio of a regular pentagon liv 10.3 onstruction of a regular pentagon with a given diagonal.... lv 10.4 Various constructions of the regular pentagon lvi 10.5 Some interesting constructions of the golden ratio lviii

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7 hapter 1 The Pythagorean theorem 1.1 The hypotenuse of a right triangle This chapter is on the famous Pythagorean theorem. We certainly all know that this is an important relation on the sides of a right triangle. The side opposite to the right angle is called the hypotenuse and is the longest among all three sides. The other two are called the legs. If we denote the legs by a and b, and the hypotenuse by c, the Pythagorean theorem says that a 2 + b 2 = c 2. c a b Why is this relation true? Suppose you have a cardboard whose length and breadth are 3 inches and 4 inches. How long is its diagonal? Your young students have not known the Pythagorean theorem yet. Help them calculate this length without throwing the Pythagorean theorem to them. Take two identical copies of the cardboard and cut each one along a diagonal. In this way there are 4 congruent right triangles. rrange these 4 triangles in this way:

8 ii The Pythagorean theorem In this figure, the 4 right triangles bound a large square outside and their hypotenuses bound another smaller square inside. We calculate areas. (1) The outer side square has side length 3+4=7units. Its area is 7 2 =49 square units. (2) The 4 right triangles make up 2 rectangular cardboards each with area 3 4 = 12 square units. The 4 triangles have total area 2 12 = 24 square units. (3) The inside square therefore has area = 25 square units. Each of its sides has length 5 units. This method indeed applies to any right triangle with given legs. Square on hypotenuse = square on sum of legs 2 rectangle with given legs as sides. b a a 3 c c 2 b b c c 4 1 a a Why is the inside region a square? b

9 1.2 The Pythagorean theorem iii 1.2 The Pythagorean theorem Rearrange the puzzle in 1.1 in another form: 2 b 4 a a 3 b 1 Here, we see the same 4 right triangles inside the same outer square. Instead of the inside square, we now have two smaller squares, each built on a leg of the right triangle. This gives the famous Pythagorean theorem: a 2 + b 2 = c Integer right triangles If we take any two numbers p and q and form then we have a =2pq, b = p 2 q 2, c = p 2 + q 2, (1.1) a 2 + b 2 = c 2. In particular, if p and q are integers, then so are a, b, and c. Sometimes, integer right triangles constructed from formula (1.1) can be reduced. For example, with p =3and q =1, we obtain (a, b, c) =(6, 8, 10), which clearly is simply the right triangle (3, 4, 5) magnified by factor 2. We say that an integer right triangle (a, b, c) is primitive if a, b, c do not have common divisor (other than 1). Every integer right triangle is a primitive one magnified by an integer factor. right triangle constructed from (1.1) is primitive if we choose the integers p, q of different parity 1 and without common divisors. 1 This means that one of them is even and the other is odd.

10 iv The Pythagorean theorem Here are all primitive integer right triangles with p, q < 10: p q a b c How many distinct integer right triangles are there whose sides are not more than 100?

11 hapter 2 The equilateral triangle Notations (O) O() O(r) circle with center O circle with center O, passing through circle with center O, radius r 2.1 onstruction of equilateral triangle n equilateral triangle is one whose three sides are equal in length. The very first proposition of Euclid s Elements teaches how to construct an equilateral triangle on a given segment. If and are the endpoints of the segments, construct the circles () and () to intersect at a point. Then, triangle is equilateral. n equilateral triangle is also equiangular. Its three angles are each equal to 60.

12 vi The equilateral triangle 2.2 The medians of an equilateral triangle Let be an equilateral triangle, and M the midpoint of. The segment M is a median of the triangle. Since M M by the test, this line M is also (i) the bisector of angle, (ii) the altitude on the side (since M ), and therefore (iii) the perpendicular bisector of the segment. M This also means that one half of an equilateral triangle is a right triangle with angles 30 and 60. We call this an right triangle. In a right triangle, the hypotenuse is twice as long as the shorter leg. The two legs are in the ratio 3:

13 2.3 The center of an equilateral triangle vii 2.3 The center of an equilateral triangle onsider two medians of the equilateral triangle intersecting at O. O N M Triangles ON and OM are both right triangles. Therefore, O =2 ON and O =2 OM. However, O + OM and O + ON have the same length. This means OM = ON, 1 and OM = 1 M, ON = 1 N. We have shown that any two medians of 3 3 an equilateral triangle intersect at a point which divides each of them in the ratio 2:1. This is the same point O for all three medians, and is called the center of the equilateral triangle. 2.4 The circumcircle and incircle of an equilateral triangle Since the three medians of an equilateral triangle have the same length, the center O is equidistant from the vertices. It is the center of the circumcircle of the equilateral triangle. The same center O is also the center of the incircle of the equilateral triangle. 1 M = N O+OM = O+ON 2 ON+OM =2 OM+ON OM = ON.

14 viii The equilateral triangle O O 2.5 Trigonometry For an acute angle θ, the trigonometric ratios sin θ, cos θ and tan θ are defined by sin θ = a c, cos θ = b c, tan θ = a b, where (a, b, c) are the sides of a right triangle containing an acute angle θ with opposite leg a. If any one of the three ratios is known, then the other two can be found. Why? (1) sin 2 θ +cos 2 θ =1. (2) tan θ = sin θ. cos θ (3) sin(90 θ) =cosθ. The precise determination of the trigonometric ratios of an angle is in general beyond the scope of elementary geometry. There are, however, a few special angles whose trigonometric ratios can be determined precisely.

15 2.5 Trigonometry ix c a θ b θ sin θ cos θ tan θ

16 x The equilateral triangle 2.6 Some interesting examples on equilateral triangles (1) For an arbitrary point P on the minor arc of the circumcircle of an equilateral triangle, P = P + P. Q Proof. If Q is the point on P such that PQ = P, then triangle PQ is equilateral since. Note that Q = P. Thus, Q P by the test. From this, Q = P, and P = Q + QP = P + P. (2) onsider a triangle whose angles are all less than 120. onstruct equilateral triangles X, Y, and Z outside the triangle. Let F be the intersection of the circumcircles of the equilateral triangles. Note that F = F = 120. It follows that F = 120, and F lies on the circumcircle of the equilateral triangle X as well. Therefore, XF = X =60, and XF + F = 180. The three points, F, X are collinear. Thus, X = F + FX = F + F + F by (1) above. Similarly,, F, Y are collinear, so do, F, Z, and Y, Z are each equal to F + F + F. Theorem 2.1 (Fermat point). If equilateral triangles X, Y, and Z are erected on the sides of triangle, then the lines X, Y, Z concur at a point F called the Fermat point of triangle. If the angles of triangle are all less than 120 (so that F is an interior point), then the segments X, Y, Z have the same length F + F + F. P

17 2.6 Some interesting examples on equilateral triangles xi Y Z F (3 ) Napoleon theorem. If equilateral triangles X, Y, and Z are erected on the sides of triangle, the centers of these equilateral triangles form another equilateral triangle. X Y Z X

18 xii The equilateral triangle

19 hapter 3 The square 3.1 onstruction of a square on a segment Given a segment, construct (1) an equilateral triangle P, (2) an equilateral triangle P Q (different from P ), (3) the segment PQand its midpoint M, (4) the ray M, (5) the circle () to intersect this ray at D, (6) the circles () and D() to intersect at. Then D is a square. Q D M P This construction gives an oriented square. If you apply it to and, you will get two different squares on the two sides of.

20 xiv The square 3.2 The diagonals of a square diagonal of a square divides it into two congruent right triangles. Each is an isosceles right triangle or a triangle. If each side of a square has unit length, a diagonal has length θ sin θ 45 1 cos θ tan θ The diagonals of a square are perpendicular to each other. They divide the square into 4 isosceles right triangles.

21 3.3 The center of the square xv 3.3 The center of the square The intersection of the diagonals is the center of the square. It is the common center of the circumcircle and the incircle of the square. 3.4 Some interesting examples on squares (1) If b a and c a are squares erected externally on the sides and of a triangle, the triangle a a has the same area as triangle. a a c b (2 ) The perpendicular from to a passes through the midpoint of. Likewise, those from to b b and from to c c pass through the midpoints of and respectively. These three perpendiculars meet at the centroid G of triangle.

22 xvi The square a a c b G b c (3 ) The midpoint of c b does not depend on the position of the vertex. It is also the same as the center of the square erected on the, on the same side as. a a c b M

23 hapter 4 Some basic principles 4.1 ngle properties Parallel lines onsider two parallel lines l 1, l 2, and a transversal L. L l 1 γ δ α l 2 β (1) The corresponding angles α and β are equal. (2) The alternate angles β and γ are equal. (3) The interior angles β and δ are supplementary, i.e., β + δ = 180. The converses of these statements are also true. In other words, if any one of (1), (2), (3) holds, then the lines l 1 and l 2 are parallel.

24 xviii Some basic principles ngle sum of a triangle The angle sum of a triangle is always 180. This is because when a side of the triangle is extended, the external angle formed is equal to the sum of the two remote internal angles. α β ngle properties of a circle (1) Let P be a point on the major arc of a circle (O). O =2 P. P P O O O Q Q P (2) Let P be a point on the minor arc of a circle (O). O +2 P = 360. (3) The opposite angles of a cyclic quadrilateral are supplementary. (4) The angle contained in a semicircle is a right angle. (5) Equal chords subtend equal angles at the center.

25 4.2 Tests of congruence of triangles xix P O O D 4.2 Tests of congruence of triangles onstruction of a triangle with three given elements triangle has six elements: three sides and three angles. onsider the construction of a triangle given three of its six elements. The triangle is unique (up to size and shape) if the given data are in one of the following patterns. (1) SSS Given three lengths a, b, c, we construct a segment with length a, and the two circles (c) and (b). These two circles intersect at two points if (and only if) b + c > a[triangle inequality]. There are two possible positions of. The resulting two triangles are congruent. (2) SS Given b, c, and angle <180, the existence and uniqueness of triangle is clear. (3) S or S These two patterns are equivalent since knowing two of the angles of a triangle, we easily determine the third (their sum being 180 ). Given, and a, there is clearly a unique triangle provided + <180. (4) RHS Given a, c and = ongruence tests These data patterns also provide the valid tests of congruence of triangles. Two triangles and XY Z are congruent if their corresponding elements are equal. Two triangles are congruent if they have three pairs of equal elements in one the five patterns above. The five valid tests of congruence of triangles are as follows.

26 xx Some basic principles (1) SSS: XYZ if = XY, = YZ, = ZX. X Z (2) SS: XY Z if Y = XY, = XY Z, = YZ. X Z Y (3) S: XYZ if = YXZ, = XY, = XY Z. X Z Y (4) S. We have noted that this is the same as S: XY Z if = YXZ, = XY Z, = YZ,.

27 4.2 Tests of congruence of triangles xxi X Z Y (5) RHS. The SS is not a valid test of congruence. Here is an example. The two triangles and XY Z are not congruent even though = YXZ, = XY, = YZ. Y However, if the equal angles are right angles, then the third pair of sides are equal: 2 = 2 2 = YZ 2 XY 2 = XZ 2, and = XZ. The two triangles are congruent by the SSS test. Without repeating these details, we shall simply refer to this as the RHS test. XY Z if = YXZ =90, = YZ, = XY. X Z Y The congruence tests for triangles form a paradigm for proofs in euclidean geometry. We shall illustrate this with numerous examples in hapter 6. Here, we give only one example, the converse of the Pythagorean theorem. Z X

28 xxii Some basic principles Theorem 4.1 (onverse of Pythagoras theorem). If the lengths of the sides of satisfy a 2 + b 2 = c 2, then the triangle has a right angle at. Y a c a b Z b X Proof. onsider a right triangle XY Z with Z =90, YZ = a, and XZ = b. y the Pythagorean theorem, XY 2 = YZ 2 + XZ 2 = a 2 + b 2 = c 2 = 2.It follows that XY =, and XYZ by the test, and = Z =90.

29 hapter 5 ircumcircle and incircle 5.1 ircumcircle The perpendicular bisector of a segment is the line perpendicular to it through its midpoint The perpendicular bisector locus theorem Theorem 5.1. point P is equidistant from and if and only if P lies on the perpendicular bisector of. P M Proof. Let M be the midpoint of. ( ) If P = P, then PM PM by the test. This means PM = PM and PM. The point P is on the perpendicular bisector of. ( ) If P is on the perpendicular bisector of, then PM PM by the test. It follows that P = P.

30 xxiv ircumcircle and incircle onstruction of circumcircle Given a triangle, there is a circle (O) through the three vertices. The center O of the circle, being equidistant from and, must lie on the perpendicular bisector of. For the same reason, it also lies on the perpendicular bisectors of and. This is called the circumcenter of triangle. O ircumcircle of a right triangle To construct a circle through the vertices of a right triangle, mark the midpoint M of the hypotenuse. The circle M() passes through and, and is the circumcircle of the right triangle. M

31 5.2 The incircle xxv 5.2 The incircle The angle bisector locus theorem The distance from a point P to a line l is the distance between P and its pedal (orthogonal projection) on l. K P H Theorem 5.2. point P is equidistant from two lines if and only if P lies on the bisector of an angle between the two lines. Proof. Let be the intersection of the two lines, and H, K the pedals of a point P on these lines. ( ) IfPH = PK then P H P K by the test. It follows that PH = PK, and P lies on the bisector of angle HK. ( ) If PH = PK, then P H P K by the test. It follows that PH = PK onstruction of incircle The incircle of a triangle is the one which is tangent to each of the three sides of the triangle. Its center is the common point of the angle bisectors, and can be located by constructing two of them. This incenter I is equidistant from the three sides. I

32 xxvi ircumcircle and incircle The incircle of a right triangle If d is the diameter of the incircle of a right triangle, then a + b = c + d. a d c b 5.3 Tangents of a circle Tangent at a point on the circle tangent to a circle is a line which intersects the circle at only one point. Given a circle O(), the tangent to a circle at is the perpendicular to the radius O at. O O M P The tangents from a point to a circle If P is a point outside a circle (O), there are two lines through P tangent to the circle. onstruct the circle with OP as diameter to intersect (O) at two points. These are the points of tangency. The two tangents have equal lengths since OP OP by the test.

33 5.4 Some interesting examples on the circumcircle and incircle of a triangle xxvii 5.4 Some interesting examples on the circumcircle and incircle of a triangle (1) Let s = 1 (a + b + c) be the semiperimeter of triangle. The lengths of 2 the tangents from the vertices to the incircle are as follows. tangent from =s a, tangent from =s b, tangent from =s c. a r r c r b (2) The following rearrangement shows that for an arbitrary triangle, the radius of its incircle is given by r = area of triangle semi-perimeter of triangle. r

34 xxviii ircumcircle and incircle (3 ) onstruct the circumcircle (O) of triangle and select an arbitrary point P on it. onstruct the reflections of P in the sidelines of the given triangle. These three reflection points always lie on a straight line l. Furthermore, as P varies on (O), the line l passes through a fixed point. P H (4 ) onstruct the circumcircle (O) and incircle (I) of a triangle. Select an arbitrary point P on (I) and construct the perpendicular to IP at P. This is a tangent to (I). Let it intersect the circle (O) at Y and Z. onstruct the tangents from Y and Z to (I). These two tangents always intersect on the circumcircle (O). Y P Z I O X

35 hapter 6 Uses of congruence tests 6.1 Isosceles triangles n isosceles triangle is one with two equal sides. Proposition 6.1. triangle is isosceles if and only if it has two equal angles. M D Proof. ( ) Let be a triangle in which =. Let M be the midpoint of. Then M M by the test. From this conclude that M = M. ( ) Let be a triangle in which =. onstruct the perpendicular from to, and let it intersect at D. Then D D by the test. From this = and the triangle is isosceles.

36 xxx Uses of congruence tests 6.2 hords of a circle (1) Let M be a point on a chord of a circle (O). M is the midpoint of if and only if OM. O M Proof. ( )IfM is the midpoint of, then OM OM by the test. Therefore, OM = OM and OM. ( ) IfOM, then OM OM by the follows that M = M, and M is the midpoint of. test. It (2) Equal chords of a circle are equidistant from the center. onversely, chords equidistant from the center are equal in length. Proof. Suppose and PQ are equal chords of a circle (O), with midpoints M and N respectively. Then M = PN, and OM OPN by the test. Therefore, OM = ON, and the chords are equidistant from the center O. onversely, if OM = ON, then OM OPN by the test. It follows that M = PN, and =2 M =2 PN = PQ.

37 6.3 Parallelograms xxxi 6.3 Parallelograms parallelogram is a quadrilateral with two pairs of parallel sides Properties of a parallelogram (1) The opposite sides of a parallelogram are equal in length. The opposite angles of a parallelogram are equal. D Proof. onsider a parallelogram D. onstruct the diagonal D. Note that D D by the test. It follows that = D and D =, and D = D. Similarly, = D. (2) The diagonals of a parallelogram bisect each other. D M Proof. onsider a parallelogram D whose diagonals and D intersect at M. Note that D = and MD M by the test. It follows that M = M and DM = M. The diagonals bisect each other.

38 xxxii Uses of congruence tests Quadrilaterals which are parallelograms (1) If a quadrilateral has two pairs of equal opposite sides, then it is a parallelogram. D Proof. Let D be a quadrilateral in which = D and D =. onstruct the diagonal D. Then D D by the test. It follows that and D//. lso, and //D. Therefore, D is a parallelogram. (2) If a quadrilateral has one pair of equal and parallel sides, then it is a parallelogram. D Proof. Suppose D is a quadrilateral in which //D and = D. Then D D by the test. It follows that D = and D is a parallelogram since Special parallelograms quadrilateral is (1) a rhombus if its sides are equal, (2) a rectangle if its angles are equal (and each 90 ), (3) a square if its sides are equal and its angles are equal.

39 6.4 The midpoint theorem and its converse xxxiii These are all parallelograms since (1) a rhombus has two pairs of sides, and (2) a rectangle has two pairs of sides. learly, then a square is a parallelogram. (1) parallelogram is a rhombus if and only if its diagonals are perpendicular to each other. (2) parallelogram is a rhombus if and only if it has an angle bisected by a diagonal. (3) parallelogram is a rectangle if and only if its diagonals are equal in length. 6.4 The midpoint theorem and its converse The midpoint theorem Given triangle, let E and F be the midpoints of and respectively. The segment FE is parallel to and its length is one half of the length of. F E D Proof. Extend FE to D such that FE = ED. Note that DE F E by the test. It follows that (i) D = F = F, (ii) DE = F E, and D//. Therefore, DF is a parallelogram, and F E//. lso, = FD =2FE.

40 xxxiv Uses of congruence tests The converse of the midpoint theorem Let F be the midpoint of the side of triangle. The parallel through F to intersects at its midpoint. F E D Proof. onstruct the parallel through to, and extend FE to intersect this parallel at D. Then, DF is a parallelogram, and D = F = F. It follows that EF ED by the test. This means that E = E, and E is the midpoint of Why are the three medians concurrent? Let E and F be the midpoints of and respectively, and G the intersection of the medians E and F. onstruct the parallel through to E, and extend G to intersect at D, and this parallel at H.. F G E D H

41 6.5 Some interesting examples xxxv y the converse of the midpoint theorem, G is the midpoint of H, and H = 2 GE Join H. y the midpoint theorem, H//F. It follows that HG is a parallelogram. Therefore, D is the midpoint of (the diagonal), and D is also a median of triangle. We have shown that the three medians of triangle intersect at G, which we call the centroid of the triangle. Furthermore, G =GH =2GD, G =H =2GE, G =H =2GF. The centroid G divides each median in the ratio 2: Some interesting examples (1) Why are the three altitudes of a triangle concurrent? Let be a given triangle. Through each vertex of the triangle we construct a line parallel to its opposite side. These three parallel lines bound a larger triangle. Note that and are both parallelograms since each has two pairs of parallel sides. It follows that = = and is the midpoint of. Y Z H X

42 xxxvi Uses of congruence tests onsider the altitude X of triangle. Seen in triangle, this line is the perpendicular bisector of since it is perpendicular to through its midpoint. Similarly, the altitudes Y and Z of triangle are perpendicular bisectors of and. s such, the three lines X, Y, Z concur at a point H. This is called the orthocenter of triangle. (2 ) The butterfly theorem. Let M be the midpoint of a chord PQ of a circle (O). and D are two chords passing through M. Join to intersect PQ at X and D to intersect PQat Y. Then MX = MY. O P Y M X Q D

43 hapter 7 The regular hexagon, octagon and dodecagon regular polygon of n sides can be constructed (with ruler and compass) if it is possible to divides the circle into n equal parts by dividing the 360 at the center into the same number of equal parts. This can be easily done for n =6(hexagon) and 8 (octagon). 1 We give some easy alternatives for the regular hexagon, octagon and dodecagon. 7.1 The regular hexagon regular hexagon can be easily constructed by successively cutting out chords of length equal to the radius of a given circle. 1 Note that if a regular n-gon can be constructed by ruler and compass, then a regular 2n-gon can also be easily constructed.

44 xxxviii The regular hexagon, octagon and dodecagon 7.2 The regular octagon (1) Successive completion of rhombi beginning with three adjacent 45 -rhombi. (2) We construct a regular octagon by cutting from each corner of a given square (of side length 2) an isosceles right triangle of (shorter) side x. This means 2 2x : x = 2:1, and 2 2x = 2x; (2 + 2)x =2, x = = 2(2 2) (2 + 2)(2 2) =2 2. D D O Q x Q x 2 2x P x P Therefore P =2 x = 2, which is half of the diagonal of the square. The point P, and the other vertices, can be easily constructed by intersecting the sides of the square with quadrants of circles with centers at the vertices of the square and passing through the center O.

45 7.3 The regular dodecagon xxxix 7.3 The regular dodecagon (1) Trisection of right angles. (2) regular dodecagon can be formed from 4 equilateral triangles inside a square. 8 of its vertices are the intersections of the sides of these equilateral triangles, while the remaining 4 are the midpoints of the sides of the squares formed by the vertices of the equilateral triangles inside the square. n easy dissection shows that the area of the regular dodecagon is 3 4 the (smaller) square containing it. of that of

46 xl The regular hexagon, octagon and dodecagon (3) Successive completion of rhombi beginning with five adjacent 30 -rhombi. This construction can be extended to other regular polygons. If an angle of measure θ := 360 can be constructed with ruler and compass, beginning with 2n n 1 adjacent θ-rhombi, by succesively completing rhombi, we obtain a regular 2n-gons tesellated by rhombi. (n 1) + (n 2) = 1 n(n 1) 2

47 hapter 8 Similar triangles 8.1 Tests for similar triangles Two triangles are similar if their corresponding angles are equal and their corresponding sides proportional to each other, i.e., XYZ if and only if = X, = Y, = Z, and YZ = ZX = XY. It is enough to conclude similarity of two triangles when they have three pairs of (1) : if the triangles have two (and hence three) pairs of equal angles. Z Y X

48 xlii Similar triangles (2) proportional SS: if ZX = XY and = X. Z Y X (3) proportional SSS: if YZ = ZX = XY. Z Y X

49 8.2 The parallel intercepts theorem xliii 8.2 The parallel intercepts theorem The intercepts on two transversals between a set of parallel lines are proportional to each other: if l 0, l 1, l 2,..., l n 1, l n are parallel lines and L 1 and L 2 are two transversals intersecting these lines at 0, 1, 2,..., n 1, n, and 0, 1, 2,..., n 1, n, then = = = n 1 n n 1 n. L 1 L 2 l l 1 l l n 1 n 1 n 1 n 1 l n n n Proof. Through 1, 2,..., n construct lines parallel to L 1, intersecting l 0, l 1,...,l n 1 at 0, 1,..., n 1 respectively. Note that (i) 1 0 = 1 0, 2 1 = 2 1,..., n n 1 = n n 1, and (ii) the triangles 0 1 0, 1 2 1,..., n 1 n n 1 are similar. It follows that 1 0 = 2 1 = = n n n 1 n Hence, = = = n 1 n n 1 n.

50 xliv Similar triangles 8.3 The angle bisector theorem The bisector of an angle divides the opposite side in the ratio of the remaining two sides, i.e.,if X bisects angle, then D X X =. Proof. onstruct the parallel through to the bisector to intersect the extension of at D. Note that D = X = X = D = D. X This means that triangle D is isosceles, and = D. parallelism of X and D, wehave Now, from the X X = D =.

51 8.4 Some interesting examples of similar triangles xlv 8.4 Some interesting examples of similar triangles (1) The altitude from the right angle vertex divides a right triangle into two triangles each similar to the given right triangle. D Proof. Let be a triangle with a right angle at. IfD is a point on such that D, then D D. Here are some interesting consequences. (a) From D D,wehave D D = D. It follows that D D 2 = D D. (b) From D,wehave D = so that 2 = D. Similarly, from D, wehave D = It follows that 2 = D. so that = D + D =(D + D) = = 2. This gives an alternative proof of the Pythagorean theorem. (a) and (b) above give simple constructions of geometric means.

52 xlvi Similar triangles onstruction 8.1. Given a point P on a segment, construct (1) a semicircle with diameter, (2) the perpendicular to at P, intersecting the semicircle at Q. Then PQ 2 = P P. Q Q θ O P θ O P (2) The radius of the semicircle is the geometric mean of P amd Q. Q O This follows from the similarity of the right triangles OP and QO. If r = O = O, then r Q = P r. From this, r2 = P Q. P

53 8.4 Some interesting examples of similar triangles xlvii (3 ) Heron s formula for the area of a triangle. Z I r I r Y s b s c Y s a The -excircle of triangle is the circle which is tangent to the side and to the extensions of and. IfY and Z are the points of tangency with and, it is easy to see that Y = Z = s, the semiperimeter. Hence, Y = s b. If the incircle touches at Y, we have known from Example 1 of 5.4 that Y = s c and Y = s a. From these we can find the radii r of the incircle, and r of the -excircle quite easily. From the similarity of triangles IY and I Y,wehave r = s a. r s Note that also IY I Y r, from which we have s b = s c. and r r r =(s b)(s c). Multiplying these two equations together, we obtain r 2 (s a)(s b)(s c) =. s From Example 2 of 5.4, the area of triangle is given by = rs. Hence, we have the famous Heron formula = s(s a)(s b)(s c).

54 xlviii Similar triangles

55 hapter 9 Tangency of circles 9.1 External and internal tangency of two circles Two circles (O) and (O ) are tangent to each other if they are tangent to a line l at the same line P, which is a common point of the circles. The tangency is internal or external according as the circles are on the same or different sides of the common tangent l. l l O O O O P P The line joining their centers passes through the point of tangency. The distance between their centers is the sum or difference of their radii, according as the tangency is external or internal.

56 l Tangency of circles 9.2 Three mutually tangent circles Given three non-collinear points,,, construct (i) triangle and its incircle, tangent to,, respectively at X, Y, Z, (ii) the circles (Z), (Y ) and (X). These are tangent to each other externally. X Y Z 9.3 Some interesting examples of tangent circles (1) In each of the following diagrams, the shaded triangle is similar to the triangle.

57 9.3 Some interesting examples of tangent circles li

58 lii Tangency of circles (2 ) Given triangle, let D, E, F be the midpoints of the sides,, respectively. The circumcircle of triangle DEF is always tangent internally to the incircle of triangle. F E I D

59 hapter 10 The regular pentagon 10.1 The golden ratio The construction of the regular pentagon is based on the division of a segment in the golden ratio. 1 Given a segment, to divide it in the golden ratio is to construct a point P on it so that the area of the square on P is the same as that of the rectangle with sides P and, i.e., P 2 = P. construction of P is shown in the second diagram. M P P 1 In Euclid s Elements, this is called division into the extreme and mean ratio.

60 liv The regular pentagon Suppose P has unit length. The length ϕ of P satisfies ϕ 2 = ϕ +1. This equation can be rearranged as ( ϕ 1 ) 2 = Since ϕ>1, wehave ϕ = 1 2 ( 5+1 ). Note that P = This explains the construction above. ϕ ϕ +1 = 1 ϕ = = The diagonal-side ratio of a regular pentagon onsider a regular pentagon DE. It is clear that the five diagonals all have equal lengths. Note that (1) = 108, (2) triangle is isosceles, and (3) = = ( ) 2=36. In fact, each diagonal makes a 36 angle with one side, and a 72 angle with another. P E D

61 10.3 onstruction of a regular pentagon with a given diagonal lv It follows that (4) triangle P is isosceles with P = P =36, (5) P = = 108, and (6) triangles and P are similar. Note that triangle P is also isosceles since (7) P = P =72. This means that P =. Now, from the similarity of and P,wehave : = : P. In other words P = P P,orP 2 = P. This means that P divides in the golden ratio onstruction of a regular pentagon with a given diagonal Given a segment, we construct a regular pentagon DE with as a diagonal. (1) Divide in the golden ratio at P. (2) onstruct the circles (P ) and P (), and let be an intersection of these two circles. (3) onstruct the circles () and () to intersect at a point D on the same side of as. (4) onstruct the circles (P ) and D(P ) to intersect at E. Then DE is a regular pentagon with as a diameter. P E D

62 lvi The regular pentagon 10.4 Various constructions of the regular pentagon (1) To construct a regular pentagon with vertices on a given circle O(), (i) construct a radius OP perpendicular to O, (ii) mark the midpoint M of OP and join it to, (iii) bisect the angle OM and let it intersect O at Q, (iv) construct the perpendicular to O at Q to intersect the circle at and E, (v) mark, D on the circle such that = and E = ED. Then DE is a regular pentagon. E Q O M P D (2) compass-only costruction of the regular pentagon: 2 construct (i) (O) =O(), (ii) (O) to mark and F on (O), (iii) (O) to mark, (iv) (O) to mark D, (v) D(O) to mark E, (vi) () and D() to intersect at X, (vii) Y = (OX) E(OX) inside (O), (viii) P = (OY ) (O), (ix) Q = P () (O), (x) D(P ) to intersect (O) at R and S (so that R is between Q and D), (xi) S(R) to intersect (O) at T. Then QRST is a regular pentagon inscribed in (O). 2 Modification of a construction given by E. P. Starke, 3 of a regular pentagon in 13 steps using compass only.

63 10.4 Various constructions of the regular pentagon lvii X Q R P D O Y S E T F (3 ) nother construction of an inscribed regular pentagon. Let O() be a given circle. (i) onstruct an isosceles triangle XY whose height is 5 of the radius of the 4 circle. (ii) onstruct (O) intersecting O() at and. (iii) Mark P = X O and Q = Y O. (iv) Join P and Q by a line intersecting the given circle at and. Then and are two sides of a regular pentagon inscribed in the circle. P Q X O Y

64 lviii The regular pentagon 10.5 Some interesting constructions of the golden ratio (1) onstruction of 36, 54, and 72 angles. Each of the following constructions begins with the division of a segment in the golden ratio at P P P 54 D 72 P (2 ) Odom s construction. Let D and E be the midpoints of the sides and of an equilateral triangle. If the line DE intersects the circumcircle of at F, then E divides DF in the golden ratio. D E F

65 10.5 Some interesting constructions of the golden ratio lix (3) Given a segment, erect a square on it, and an adjacent one with base. IfD is the vertex above, construct the bisector of angle D to intersect at P. Then P divides in the golden ratio. D Proof. If =1, then P P : P =1 : 5, P : =1 : 5+1 :2 =1 : 5+1 : P = 5+1:2 =ϕ :1. This means that P divides in the golden ratio. (4) Hofstetter s parsimonious construction. 4 Given a segment, construct (i) 1 = (), (ii) 2 = (), intersecting 1 at and D, (iii) the line to intersect 1 at E (apart from ), (iv) 3 = E() to intersect 2 at F (so that and F are on opposite sides of ), (v) the segment F to intersect at G. Then the point G divides the segment in the golden section. 4 K. Hofstetter, nother 5-step division of a segment in the golden section, Forum Geom., 4 (2004)

66 lx The regular pentagon 3 D F 2 1 E G (5) Hofstetter s rusty compass construction. 5 (i) onstruct () and () intersecting at and D. (ii) Join D to intersect at its midpoint M. (iii) onstruct M() to intersect () at E on the same side of as. (iv) Join DE to intersect at P. Then P divides the segment in the golden ratio. E M P D 5 K. Hofstetter, Division of a segment into golden ratio with ruler and rusty compass, Forum Geom., 5 (2005)