hapter 1 Some asic Theorems 1.1 The ythagorean Theorem Theorem 1.1 (ythagoras). The lengths a b < c of the sides of a right triangle satisfy the relation a 2 + b 2 = c 2. roof. b a a 3 2 b 2 b 4 b a b 4 1 a a 3 1 a b Theorem 1.2 (onverse of ythagoras theorem). If the lengths of the sides of satisfy a 2 + b 2 = c 2, then the triangle has a right angle at.
102 Some asic Theorems Y a c a b Z b roof. onsider a right triangle Y Z with Z = 90, Y Z = a, and Z = b. y the ythagorean theorem, Y 2 = Y Z 2 + Z 2 = a 2 + b 2 = c 2 = 2. It follows that Y =, and Y Z by the SSS test, and = Z = 90.
1.1 The ythagorean Theorem 103 Exercise. 1. right triangle has shorter sides a and b. The altitude on the hypotenuse has length d. Show that 1 a 2 + 1 b 2 = 1 d 2. a d b 2. Let and E be the midpoints of the sides and of an equilateral triangle. If the line E intersects the circumcircle of at F, calculate the ratio E : EF. E F 3. Given a segment, erect a square on it, and an adjacent one with base. If is the vertex above, construct the bisector of angle to intersect at. alculate the ratio :.
104 Some asic Theorems 4. semicircle with diameter is constructed outside an equilateral triangle. and Y are points dividing the semicircle into three equal parts. Show that the lines and Y divide the side into three equal parts. Y 5. is a chord of length 2 in a circle (2). is the midpoint of the minor arc and M the midpoint of the chord. M Show that (i) M = 2 3; (ii) = 6 2. educe that tan15 = 2 3, sin 15 = 1 4 ( 6 2), cos 15 = 1 4 ( 6+ 2).
1.1 The ythagorean Theorem 105 6. is a square erected externally on the hypotenuse of the right triangle. If, intersect at, Y respectively, show that Y 2 = Y. Y 7. is a rectangle erected externally on the hypotenuse of the right triangle, with = 2. If, intersect at, Y respectively, show that 2 + Y 2 = 2. Y
106 Some asic Theorems 8. In the diagram below, = and = 60. rove that = 3. 60 9. In the diagram below, = and = 30. rove that = 3. 30
1.1 The ythagorean Theorem 107 10. and Y are equilateral triangles inside a rectangle. The lines and Y are extended to intersect and respectively at and. Show that (a) is an equilateral triangle; (b) + =. Y
108 Some asic Theorems Excursus: onstructions of geometric mean We present two ruler-and-compass constructions of the geometric means of two quantities given as lengths of segments. These are based on Euclid s proof of the ythagorean theorem. onstruct the altitude at the right angle to meet at and the opposite side ZZ of the square ZZ at. Note that the area of the rectangle Z is twice of the area of triangle Z. y rotating this triangle about through a right angle, we obtain the congruent triangle, whose area is half of the area of the square on. It follows that the area of rectangle Z is equal to the area of the square on. For the same reason, the area of rectangle Z is equal to that of the square on. From these, the area of the square on is equal to the sum of the areas of the squares on and. Y Y Z Z
1.1 The ythagorean Theorem 109 onstruction 1.1. Given two segments of length a < b, mark three points,, on a line such that = a, = b, and, are on the same side of. escribe a semicircle with as diameter, and let the perpendicular through intersect the semicircle at. Then 2 =, so that the length of is the geometric mean of a and b. ab a b onstruction 1.2. Given two segments of length a, b, mark three points,, on a line ( between and ) such that = a, = b. escribe a semicircle with as diameter, and let the perpendicular through intersect the semicircle at. Then 2 =, so that the length of is the geometric mean of a and b. ab a b
110 Some asic Theorems 1.2 Some theorems on circles 1.2.1 The perpendicular bisector locus theorem The perpendicular bisector of a segment is the line perpendicular to it through its midpoint. Theorem 1.3. point is equidistant from and if and only if lies on the perpendicular bisector of. M roof. Let M be the midpoint of. ( ) If =, then M M by the SSS test. This means M = M and M. The point is on the perpendicular bisector of. ( ) If is on the perpendicular bisector of, then M M by the SS test. It follows that =. The perpendicular bisectors of the three sides of a triangle are concurrent at the circumcenter of the triangle. This is the center of the circumcircle, the circle passing through the three vertices of the triangle. The circumcenter of a right triangle is the midpoint of its hypotenuse.
1.2 Some theorems on circles 111 1.2.2 The intersecting chords theorem Theorem 1.4. Given a point and a circle (r), if a line through intersects the circle at two points and, then = 2 r 2, independent of the line. M M roof. Let M be the midpoint of. Note that M is perpendicular to. If is in the interior of the circle, then = (M + M)(M M) = (M + M)(M M) = M 2 M 2 = (M 2 + M 2 ) (M 2 + M 2 ) = r 2 2. The same calculation applies to the case when is outside the circle, provided that the lengths of the directed segments are signed. The product r 2 2 is called the power of with respect to the circle. It is positive, zero, or negative according as is inside, on, or outside the circle.
112 Some asic Theorems orollary 1.5 (Intersecting chords theorem). If two chords and of a circle intersect, extended if necessary, at a point, then =. In particular, if the tangent at T intersects at, then = T 2. T The converse of the intersecting chords theorem is also true. Theorem 1.6. Given four points,,,, if the lines and intersect at a point such that = (as signed products), then,,, are concyclic. In particular, if is a point on a line, and T is a point outside the line such that = T 2, then T is tangent to the circle through,, T. 1.2.3 onstruction of tangents of a circle tangent to a circle is a line which intersects the circle at only one point. Given a circle (), the tangent to a circle at is the perpendicular to the radius at. If is a point outside a circle (), there are two lines through tangent to the circle. onstruct the circle with as diameter to intersect () at two points. These are the points of tangency. The two tangents have equal lengths since the triangles and are congruent by the RHS test. 1.2.4 enters of similitude of two circles Given two circles (R) and I(r), whose centers and I are at a distance d apart, we animate a point on (R) and construct a ray through I oppositely parallel
1.2 Some theorems on circles 113 M to the ray to intersect the circle I(r) at a point Y. The line joining and Y intersects the line I of centers at a point T which satisfies T : IT = : IY = R : r. This point T is independent of the choice of. It is called the internal center of similitude, or simply the insimilicenter, of the two circles. Y T T Y I If, on the other hand, we construct a ray through I directly parallel to the ray to intersect the circle I(r) at Y, the line Y always intersects I at another point T. This is the external center of similitude, or simply the exsimilicenter, of the two circles. It divides the segment I in the ratio T : T I = R : r. 1.2.5 External and internal tangency of two circles Two circles () and ( ) are tangent to each other if they are tangent to a line l at the same line, which is a common point of the circles. The tangency is
114 Some asic Theorems internal or external according as the circles are on the same or different sides of the common tangent l. T I I T 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.
1.2 Some theorems on circles 115 Exercise. 1. In the diagram,,, Y, and are squares. Show that,, Z, are concyclic. Y Z 2. Show that T = RS. T S R 3. Show that the tangent to the semicircle bounds a (3, 4, 5) triangle with two sides of the square.
116 Some asic Theorems 4. circle is inscribed in a square. second circle on diameter E touches the first circle. Show that = 4 E. E 5. alculate the ratio :. M 6. alculate the ratio :.
1.2 Some theorems on circles 117 7. Show that the two segments joining pairs of centers are perpendicular and equal in length, and compute the ratio :. 8. Show that the centers of three circles form an equilateral triangle, and compute the ratio :. 1 9. From the centers of each of two nonintersecting circles tangents are drawn to the other circle. rove that the chords and Y are equal in length. Y 1 3 : 3 + 2.
118 Some asic Theorems 10. The centers and of two circles (a) and (b) are at a distance d apart. The line intersect the circles at and respectively, so that, are between,. 11. Two congruent circles of radii a have their centers on each other. alculate the radius of the circle tangent to one of them internally, the other externally, and the line joining their centers. 12. (a) and (b) are two semicircles tangent internally to each other at. circle K(r) is constructed tangent externally to (a), internally to (b), and to the line at a point. Show that = b(3a b), and r = a + b 4ab(b a) (a + b) 2. K
1.2 Some theorems on circles 119 Excursus: tolemy s theorem Theorem 1.7 (tolemy). convex quadrilateral is cyclic if and only if + =. roof. (Necessity) ssume, without loss of generality, that >. hoose a point on the diagonal such that =. Triangles and are similar, since =. It follows that : = :, and =. Now, triangles and are also similar, since = + = + =, and =. It follows that : = :, and ombining the two equations, we have =. + = ( + ) =.
120 Some asic Theorems (Sufficiency). Let be a quadrilateral satisfying (**). Locate a point such that = and =. Then the triangles and are similar. It follows that : : = : :. d From this we conclude that (i) =, and (ii) triangles and are similar since = and : = :. onsequently, : = :, and ombining the two equations, =. ( + ) = cot + =. It follows that + =, and the point lies on diagonal. From this, = =, and the points,,, are concyclic.
1.2 Some theorems on circles 121 Exercise. 1. is an equilateral triangle inscribed in a circle. is a point on the minor arc. The line intersects at. Show that (i) = +, 1 (ii) = 1 + 1, (iii) 2 + 2 + 2 is constant. 2. Each diagonal of a convex quadrilateral bisects one angle and trisects the opposite angle. etermine the angles of the quadrilateral. 2 3. (a) If three consecutive sides of a convex, cyclic quadrilateral have lengths a, b, c, and the fourth side d is a diameter of the circumcircle, show that d is the real root of the cubic equation x 3 (a 2 + b 2 + c 2 )x 2abc = 0. b a c x (b) alculate the diameter if a = 2, b = 7 and c = 11. 2 nswer: Either = = 72, = = 108, or = = 720 7, = = 540 7.
122 Some asic Theorems Excursus: The butterfly theorem Theorem 1.8. Let M be the midpoint of a chord of a circle (). Two chords HK and are drawn through M. Let H and K intersect at and Y respectively. Then M is the midpoint of Y. H M Y K roof. onstruct the circle (M) to intersect and HK at E and F respectively. Extend EF to intersect H and K at and G. y symmetry, HM = FK and E = M. H M Y E F G K
1.2 Some theorems on circles 123 Since is tangent to the circle (M), we easily note the similarity of the following pairs of triangles: E and KY M, EG and HM, HF and Y M, KFG and M. Note also that G = E = KY M = GY. It follows that the points,, Y, G are concyclic. We show that the center of the circle containing them coincides with. Then = Y, and M = MY. Making use of the similarity of triangle pairs in (i) and (ii), we have E EG = E Y M Y M EG M M = E KM E HM Y M M = M KM M HM Y M M M M = KM HM Y M M M M = HM MK Y M M = Y M M. lso, making use of similarity of triangle pairs in (iii) and (iv), we have F FG = F Y M Y M FG M M = HF M FK M Y M M = MK M HM M Y M M HM MK = M M Y M M = Y M M. This means that the points E, F, M have equal powers in the circle through,, Y, G. ut these points have equal powers in the given circle. These two circles have the same center. We have proved that = Y. From this, M is the midpoint of Y.
124 Some asic Theorems 1.3 The regular pentagon 1.3.1 The golden ratio Given a segment, a point in the segment is said to divide it in the golden ratio if 2 =. Equivalently, = 5+1. We shall denote this golden 2 ratio by ϕ. It is the positive root of the quadratic equation x 2 = x + 1. M onstruction 1.3 (ivision of a segment in the golden ratio). Given a segment, (1) draw a right triangle M with M perpendicular to and half in length, (2) mark a point on the hypotenuse M such that M = M, (3) mark a point on the segment such that =. Then divides into the golden ratio.
1.3 The regular pentagon 125 1.3.2 The regular pentagon onsider a regular pentagon E. It is clear that the five diagonals all have equal lengths. Note that (1) = 108, (2) triangle is isosceles, and (3) = = (180 108 ) 2 = 36. In fact, each diagonal makes a 36 angle with one side, and a 72 angle with another. E It follows that (4) triangle is isosceles with = = 36, (5) = 180 2 36 = 108, and (6) triangles and are similar. Note that triangle is also isosceles since (7) = = 72. This means that =. Now, from the similarity of and, we have : = :. In other words =, or 2 =. This means that divides in the golden ratio. onstruction 1.4. Given a segment, we construct a regular pentagon E with as a diagonal. (1) ivide in the golden ratio at. (2) onstruct the circles () and (), and let be an intersection of these two circles. (3) onstruct the circles () and () to intersect at a point on the same side of as. (4) onstruct the circles () and () to intersect at E. Then E is a regular pentagon with as a diagonal.
126 Some asic Theorems 1.3.3 onstruction of 36, 54, and 72 angles ngles of sizes 36, 54, and 72 can be easily constructed from a segment divided in the golden ratio. 36 36 54 36 54 72 72
1.3 The regular pentagon 127 Exercise. 1. is a square inscribed in a semicircle with diameter Y. Show that divides Y in the golden ratio. Y 2. The two congruent circles are tangent to each other. Suppose divides in the golden ration,, and. Show that Y divides in the golden ratio. Y 3. is an isosceles triangle with a point on such that = =. Show that (i) = 36, and (ii) : = ϕ : 1. Suppose = 1. Let E be the midpoint of the side. Show that E = 1 10 + 2 5. 4 educe that cos 36 = 5 + 1, sin36 = 1 10 2 5, tan 36 = 5 2 5. 4 2
128 Some asic Theorems 4. is an isosceles triangle with = = 4. is a point on such that = =. Let be the midpoint of. alculate the length of, and deduce that sin18 = 5 1, cos 18 = 1 10 + 2 5, tan 18 = 1 25 10 5. 4 4 5 5. Justify the following construction of the regular pentagon. Let and Y be two perpendicular radii of a circle, center. Y M E (a) Mark the midpoint M of Y and bisect angle M to intersect at. (b) onstruct the perpendicular to at to intersect the circle at and E. Then,, E are three adjacent vertices of a regular pentagon inscribed in the circle. The remaining two vertices can be easily constructed. 6. If the legs and the altitude of a right triangle form the sides of another right triangle, show that the altitude divides the hypotenuse into the golden ratio.
1.3 The regular pentagon 129 Excursus: roof without words 1 + 2 sin18 = 2 sin54 = 1 2 sin18. 36 36 1 1 2 sin54 = 1 2sin 18 54 18 72 72 54 2 sin18 = ϕ 1 1 18