Theorem 1.2 (Converse of Pythagoras theorem). If the lengths of the sides of ABC satisfy a 2 + b 2 = c 2, then the triangle has a right angle at C.


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1 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 + b = c. roof. b a a 3 b b 4 b a b 4 1 a a 3 1 a b Theorem 1. (onverse of ythagoras theorem). If the lengths of the sides of satisfy a + b = c, then the triangle has a right angle at. a c a b Z b roof. onsider a right triangle Z with Z = 90, Z = a, and Z = b. y the ythagorean theorem, = Z + Z = a + b = c =. It follows that =, and Z by the SSS test, and = Z = 90.
2 10 Some asic Theorems Example. Trigonometric ratios of 30, 45, and 60 : θ sinθ cos θ tanθ Example. Given triangle with a fixed base, squares b a and c a are constructed externally. The midpoint of the segment c b does not depend on the position of the vertex. a a M c b Z
3 1.1 The ythagorean Theorem 103 roof. Let be the midpoint of, and,, Z the pedals of, b and c on the line. There are two pairs of congruent triangles, b and c Z. Note that is the midpoint of Z. If M is the midpoint of c b, then M = 1 ( b + Z c ) = 1 ( + ) = 1. This means that M is the center of the square on, on the same side of. Example. For an arbitrary point on the minor arc of the circumcircle of an equilateral triangle, = +. roof. If is the point on such that =, then the isosceles triangle is equilateral since = = 60. Note that =. Thus, by the SS congruence test. From this, =, and = + = +.
4 104 Some asic Theorems Exercise 1. 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. triangle has =. Show that b = c(a + c). istinguish between the cases when < 90, = 90, and > 90. c c b c b M c c M a 3. The diagonals of a square intersect at. The bisector of angle intersects at E and at F. rove that (i) E = F, (ii) E = F. 1 E F 1 Hint: onstruct the parallel through to, to intersect E at G.
5 1.1 The ythagorean Theorem is a right triangle with = 30 and = 90, and = 1. The side is extended to such that =. (a) Show that = 6 +. From this deduce that tan15 = 3, sin 15 = 6, cos 15 = (b) The triangle is reflected about the hypotenuse so that the image of intersects at. Show that (i) =, (ii) = 6 (Hint: onsider the pedal Z of on the line E). E Z (c) Let the perpendicular to at intersect at F. Show that F = 4 and F = 6 (Hint: Extend to G such that G = F ). E G F
6 106 Some asic Theorems 5. In the diagram below, = and = 60. rove that = In the diagram below, = and = 30. rove that = and are equilateral triangles inside a rectangle. The lines and are extended to intersect and respectively at and. Show that (a) is an equilateral triangle; (b) + =.
7 1.1 The ythagorean Theorem is an isosceles triangle with = = 80. and E are points on and respectively such that = 60 and E = 50. alculate E. E Hint: Let F be the point on such that F =. Show that F = EF = F.
8 108 Some asic Theorems onstructions of geometric mean We present two rulerandcompass 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. Z Z 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 =, so that the length of is the geometric mean of a and b. ab a b
9 1.1 The ythagorean Theorem 109 onstruction 1.. 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 =, so that the length of is the geometric mean of a and b. ab a b Exercise 1. alculate the ratio :. Hint: onsider the pedal of on and note that : = :. M
10 110 Some asic Theorems. is a square erected externally on the hypotenuse of the right triangle. If, intersect at, respectively, show that =. 3. is a rectangle erected externally on the hypotenuse of the right triangle, with =. If, intersect at, respectively, show that + =.
11 1. The golden ratio The golden ratio Given a segment, a point in the segment is said to divide it in the golden ratio if =. Equivalently, = 5+1. We shall denote this golden ratio by ϕ. It is the positive root of the quadratic equation x = 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, () 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. Suppose has unit length. The length ϕ of satisfies ϕ = ϕ + 1. This equation can be rearranged as Since ϕ > 1, we have ( ϕ 1 ) = 5 4. ϕ = 1 ( ). Note that = This explains the construction above The regular pentagon ϕ ϕ + 1 = 1 ϕ = = onsider a regular pentagon E. It is clear that the five diagonals all have equal lengths. Note that (1) = 108,
12 11 Some asic Theorems E () triangle is isosceles, and (3) = = ( ) = 36. In fact, each diagonal makes a 36 angle with one side, and a 7 angle with another. It follows that (4) triangle is isosceles with = = 36, (5) = = 108, and (6) triangles and are similar. Note that triangle is also isosceles since (7) = = 7. This means that =. Now, from the similarity of and, we have : = :. In other words =, or =. 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. () 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.
13 1. The golden ratio onstruction of 36, 54, and 7 angles ngles of sizes 36, 54, and 7 can be easily constructed from a segment divided in the golden ratio ϕ cos 36 = ϕ ϕ cos 7 = 1 ϕ.
14 114 Some asic Theorems Exercise 1. Three equal segments 1 1,, 3 3 are positioned in such a way that the endpoints, 3 are the midpoints of 1 1, respectively, while the endpoints 1,, 3 are on a line perpendicular to 1 1. Show that divides 1 3 in the golden ratio Given an equilateral triangle, erect a square E externally on the side. onstruct the circle, center, passing through E, to intersect the line at F. Show that divides F in the golden ratio. E 3. If the legs and the altitude of a right triangle form the sides of another right triangle, show that the altitude divides the hypotenuse in the golden ratio. F 36
15 1. The golden ratio is a square inscribed in a semicircle with diameter. Show that divides in the golden ratio. 5. The two congruent circles are tangent to each other. Suppose divides in the golden ration,, and. Show that divides in the golden ratio. 6. Justify the following construction of the regular pentagon. Let and be two perpendicular radii of a circle, center. M E (a) Mark the midpoint M of 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.
16 116 Some asic Theorems 1.3 asic construction principles (1) erpendicular bisector locus variable point is equidistant from two fixed points and if and only if lies on the perpendicular bisector of the segment. 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. () ngle bisector locus variable point is equidistant from two fixed lines l and l if and only if lies on the bisector of one of the angles between l and l. (3) Tangency of circles Two circles are tangent to each other if and only if they have a common point on the line joining their centers.
17 1.3 asic construction principles 117 Example. Given two congruent circles each with center on the other circle, to construct a circle tangent to the center line, and also to the given circles, one internally and the other externally. K T x Let = a. Suppose the required circle has radius r, and T = x, where T is the point of tangency T with the center line. From these, we have (a + x) + r = (a + r), x + r = (a r). x + a = 3 a, a + x = r. This means that if M is the midpoint of, then MT = 3 a, which is the height of the equilateral triangle on. In other words, if is an intersection of the two given circles, then M and MT are two adjacent sides of a square. Furthermore, the side opposite to M is a diameter of the required circle! K 3a T x a M
18 118 Some asic Theorems 1.4 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. M 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. Exercise 1. Show that T = RS. T S R. Show that the tangent to the semicircle bounds a (3, 4, 5) triangle with two sides of the square.
19 1.4 onstruction of tangents of a circle From the centers of each of two nonintersecting circles tangents are drawn to the other circle. rove that the chords and are equal in length. 4. 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,. From the extremity tangents are constructed to the circle (b). alculate the radius of the circle tangent internally to (a) and to these tangent lines. Show that this is congruent to the circle tangent internally to (b) and to the tangents from to (a).
20 10 Some asic Theorems 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 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. Exercise 1. circle is inscribed in a square. second circle on diameter E touches the first circle. Show that = 4 E. E. alculate the ratio :.
21 1.4 onstruction of tangents of a circle Show that the two segments joining pairs of centers are perpendicular and equal in length, and compute the ratio :. 4. Show that the centers of three circles form an equilateral triangle, and compute the ratio : 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 to the ray to intersect the circle I(r) at a point. The line joining and intersects the line I of centers at a point T which satisfies T : IT = : I = 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. T T I If, on the other hand, we construct a ray through I directly parallel to the ray to intersect the circle I(r) at, the line 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. 3 + : 3.
22 1 Some asic Theorems 1.5 The intersecting chords theorem Theorem 1.3. Given a point and a circle (r), if a line through intersects the circle at two points and, then = r, 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 M = (M + M ) (M + M ) = r. 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 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. orollary 1.4 (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. T The converse of the intersecting chords theorem is also true.
23 1.5 The intersecting chords theorem 13 Theorem 1.5. 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, then T is tangent to the circle through,, T. Example. Let be a triangle with =. Then b = c(c + a). E F onstruct a parallel through to the bisector E, to intersect the extension of at F. Then F = E = 1 =. This means that is tangent to the circle through,, F. y the intersecting chord theorem, = F, i.e., b = c(c + a). Exercise 1. Let be an interior point of an equilateral triangle such that = 10. The lines and intersect and at and Z respectively. Show that : Z = :. Z
24 14 Some asic Theorems 1.6 Radical axis Given two nonconcentric circles 1 and. The locus of points of equal powers with respect to the circle is a straight line perpendicular to the line joining their centers. In fact, if the circles are concentric, there is no finite point with equal powers with respect to the circles. n the other hand, if the centers are distinct points and at a distance d apart, there is a unique point with distances = x and = d x such that r 1 x = r (d x). If this common value is m, then every point on the perpendicular to at has power m with respect to each of the circles. This line is called the radical axis of the two circles. If the two circles intersect at two distinct points, then the radical axis is the line joining these common points. If the circles are tangent to each other, then the radical axis is the common tangent. Let 1 and be the centers of 1 and, with midpoint M. If and are on the same side of the line 1, with 1 and perpendicular to 1, construct the perpendicular bisector of to intersect 1 at H. Let K be the reflection of K in M, the perpendicular at K to 1 is the radical axis of the circles 1 and. H 1 K M Here is another construction. If 1, 1 are the tangents from 1 to, and, those from to, the lines and are parallel. The line equidistant from and is the radical axis. 1
25 1.6 Radical axis 15 Theorem 1.6. Given three circles with distinct centers, the radical axes of the three pairs of circles are either concurrent or are parallel. roof. (1) If any two of the circles are concentric, there is no finite point with equal powers with respect to the three circles. () If the centers of the circles are distinct and noncollinear, then two of the radical axes, being perpendiculars to two distinct lines with a common point, intersect at a point. This intersection has equal powers with respect to all three circles, and also lies on the third radical axis. (3) If the three centers are distinct but collinear, then the three radical axes three parallel lines, which coincide if any two of them do. This is the case if and only if the three circles two points in common, or at mutually tangent at a point. In this case we say that the circles are coaxial. If the three circles have noncollinear centers, the unique point with equal powers with respect to the circles is called the radical center. Example. Let 1, be points on the sideline, 1, on, and Z 1, Z on of triangle. If 1 = Z 1 Z, Z 1 Z = 1, 1 = 1, then the six points 1,, 1,, Z 1, Z are concyclic. roof. y the intersecting chords theorem, the points 1,, Z 1, Z lie on a circle 1. Likewise, Z 1, Z, 1, lie on a circle, and 1,, 1, lie on a circle 3. If any two of these circles coincide, then all three circles coincide. If the circles are all distinct, then the sidelines of the triangle, being the three radical axes of the circles and nonparallel, should concur at a point, a contradiction. Exercise 1. Let, E, F be the midpoints of the sides,, of triangle, and,, Z the pedals of,, on their opposite sides. rove that the six points, E, F,,, Z are concyclic.
26 16 Some asic Theorems 1.7 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 + = ( + ) =. d (Sufficiency). Let be a quadrilateral satisfying (**). Locate a point such that = and =. Then the triangles and are similar. It follows that : : = : :. From this we conclude that (i) =, and (ii) triangles and are similar since = and : = :. onsequently, : = :, and ombining the two equations, =. ( + ) = + =. It follows that + =, and the point lies on diagonal. From this, = =, and the points,,, are concyclic.
27 1.7 tolemy s theorem 17 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) + + is constant.. Each diagonal of a convex quadrilateral bisects one angle and trisects the opposite angle. etermine the angles of the quadrilateral (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 + b + c )x abc = 0. b a c x (b) alculate the diameter if a =, b = 7 and c = nswer. Either = = 7, = = 108, or = = 70 7, = =
28 18 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 respectively. Then M is the midpoint of. H M 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 E M F G K Since is tangent to the circle (M), we easily note the similarity of the following pairs of triangles: E and K M, EG and HM, HF and M, KFG and M. Note also that G = E = K M = G. It follows that the points,,, G are concyclic. We show that the center of the circle containing them coincides with. Then =, and M = M.
29 1.7 tolemy s theorem 19 Making use of the similarity of triangle pairs in (i) and (ii), we have E EG = E M M EG M M = E KM E HM M M = M KM M HM M M M M = KM HM M M M M = HM MK M M = M M. lso, making use of similarity of triangle pairs in (iii) and (iv), we have F FG = F M M FG M M = HF M FK M M M = MK M HM M M M HM MK = M M M M = M M. This means that the points E, F, M have equal powers in the circle through,,, G. ut these points have equal powers in the given circle. These two circles have the same center. We have proved that =. From this, M is the midpoint of.
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