hapter 2 The laws of sines and cosines 2.1 The law of sines Theorem 2.1 (The law of sines). Let R denote the circumradius of a triangle. 2R = a sin α = b sin β = c sin γ. α O O α as Since the area of a triangle is given by = 1 bc sin α, the circumradius can be written 2 R = abc 4.
132 The laws of sines and cosines xercise 1. In triangle, = 60. If and are points on and such that bisects angle and bisects angle, then = 30. I 2. In triangle, = 120. If and are points on and such that bisects angle and bisects angle, then = 30. int: onstruct the circle to intersect at. I 3. Given triangle, let Y be the intersection of the perpendicular bisector of and the perpendicular to at. Similarly, let Z be the intersection of the perpendicular bisector of and the perpendicular to at. Show that Y Z = R 2, where R is the circumradius of triangle. Y Z
2.1 The law of sines 133 4. is a triangle with = 60. The intersection P of the perpendicular bisector of and a trisector of angle is such that P =. alculate angles and. P 5. Inside triangle there is a point P such that P = 20, P = 10, P = 40, and P = 30. Show that triangle is isosceles. 20 P 10 40 30 6. In the diagram below, the angles, and are equal. (a) Show that triangles and are similar. (b) Show that the ratio of similarity is = cos + ( 1 + cosαcosβ cosγ sin α sin β sin γ ) sin.
134 The laws of sines and cosines 2.2 The orthocenter 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 X 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. This is called the orthocenter of triangle. Proposition 2.1. The reflections of the orthocenter in the sidelines lie on the circumcircle. O a Proof. It is enough to show that the reflection a of in lies on the circumcircle. onsider also the reflection O a of O in. Since and OO a are parallel and have the same length (2R cosα), OO a is a parallelogram. On the other hand, OO a a is a isosceles trapezoid. It follows that O a = O a = O, and a lies on the circumcircle. O a
2.2 The orthocenter 135 xercise 1. triangle is equilateral if and only if its circumcenter and centroid coincide. 2. Let be the orthocenter of triangle. Show that (i) is the orthocenter of triangle ; (ii) the triangles,, and have the same circumradius. 3. In triangle with circumcenter O, orthocenter, midpoint of, and perpendicular foot X of on, OX is a rectangle of dimensions 11 5. alculate the length of the side. 11 O 5 X 4. In triangle, one pair of trisectors of the angles and meet at the orthocenter. (a) Show that the other pair of trisectors of these angles meet at the circumcenter. (b) ind all possible values of the angles of the triangle. O O O
136 The laws of sines and cosines 2.3 The law of cosines Given a triangle, we denote by a, b, c the lengths of the sides,, respectively. Theorem 2.2 (The law of cosines). c 2 = a 2 + b 2 2ab cosγ. c b c b X a a X Proof. Let X be the altitude on. c 2 = X 2 + X 2 = (a b cosγ) 2 + (b sin γ) 2 = a 2 2ab cos γ + b 2 (cos 2 γ + sin 2 γ) = a 2 + b 2 2ab cosγ. 2.3.1 Stewart s theorem Theorem 2.3 (Stewart). Let X be a point on the sideline of triangle. a X 2 = X b 2 + X c 2 a X X. ere, the lengths of the directed segments on the line are signed. quivalently, if X : X = λ : µ, then X 2 = λb2 + µc 2 λ + µ λµa2 (λ + µ) 2. Proof. Use the cosine formula to compute the cosines of the angles X and X, and note that cos X = cos X.
2.3 The law of cosines 137 2.3.2 Napoleon s theorem If similar isosceles triangles X, Y and Z (of base angle ) are constructed externally on the sides of triangle, the lengths of the segments Y Z, ZX, XZ can be computed easily. or example, in triangle Y Z, Y = b sec, Z = c sec and 2 2 Y Z = α + 2. Y Z X y the law of cosines, Likewise, we have Y Z 2 = Y 2 + Z 2 2Y Z cosy Z = sec2 (b 2 + c 2 2bc cos(α + 2)) 4 = sec2 (b 2 + c 2 2bc cosαcos 2 + 2bc sin α sin 2) 4 = sec2 (b 2 + c 2 (b 2 + c 2 a 2 ) cos 2 + 4 sin 2) 4 = sec2 (a 2 cos 2 + (b 2 + c 2 )(1 cos 2) + 4 sin 2). 4 ZX 2 = sec2 (b 2 cos 2 + (c 2 + a 2 )(1 cos 2) + 4 sin 2), 4 XY 2 = sec2 (c 2 cos 2 + (a 2 + b 2 )(1 cos 2) + 4 sin 2). 4 It is easy to note that Y Z = ZX = XY if and only if cos 2 = 1 2, i.e., = 30. In this case, the points X, Y, Z are the centers of equilateral triangles erected externally on,, respectively. The same conclusion holds if the equilateral triangles are constructed internally on the sides. This is the famous Napoleon theorem. Theorem 2.4 (Napoleon). If equilateral triangles are constructed on the sides of a triangle, either all externally or all internally, then their centers are the vertices of an equilateral triangle.
138 The laws of sines and cosines Y Z Z X Y X xample. Given three points,, that form an acute-angled triangle, construct three circles with these points as centers that are mutually orthogonal to each other. Y Z X Let = a, = b, and = c. If these circles have radii R a, R b, R c respectively, then R 2 b + R2 c = a2, R 2 c + R2 a = b2, R 2 a + R2 b = c2. rom these, R 2 a = 1 2 (b2 + c 2 a 2 ), R 2 b = 1 2 (c2 + a 2 b 2 ), R 2 c = 1 2 (a2 + b 2 c 2 ). These are all positive since is an acute triangle. onsider the perpendicular foot of on. Note that = c cos, so that R 2 a = 1 2 (b2 +c 2 a 2 ) = bc cos =. It
2.3 The law of cosines 139 follows if we extend to intersect at Y the semicircle constructed externally on the side as diameter, then, Y 2 = = R 2 a. Therefore we have the following simple construction of these circles. (1) With each side as diameter, construct a semicircle externally of the triangle. (2) xtend the altitudes of the triangle to intersect the semicircles on the same side. Label these X, Y, Z on the semicircles on,, respectively. These satisfy Y = Z, Z = X, and X = Y. (3) The circles (Y ), (Z) and (X) are mutually orthogonal to each other.
140 The laws of sines and cosines xercise 1. is a triangle with = 7, = 7 + 1, and = 7 1. Show that α = 90 + β and find cosγ. 7 7 1 7 + 1 2. In the diagram below, show that T 1 + T 2 + T 3 = 3(S 1 + S 2 + S 3 ). T 3 S 2 S 1 T 2 S 3 T 1 3. is a triangle with a = 12, b + c = 18, and cos = 7. alculate the lengths of 38 b and c, 1 and show that a 3 = b 3 + c 3. 4. Given triangle, points X, Y, Z are chosen on the extensions of the half-lines,, respectively such that Y = Z = X. Show that triangle XY Z is equilateral if and only if triangle is equilateral. Z γ α β Y X 1 MM 688, P.. Pizá. ere, b = 9 5, and c = 9 + 5.
2.4 The medians 141 2.4 The medians 2.4.1 The centroid Let and be the midpoints of and respectively, and G the intersection of the medians and. onstruct the parallel through to, and extend G to intersect at, and this parallel at.. G y the converse of the midpoint theorem, G is the midpoint of, and = 2 G Join. y the midpoint theorem, //. It follows that G is a parallelogram. Therefore, is the midpoint of (the diagonal), and 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. urthermore, G =G = 2G, G = = 2G, G = = 2G. The centroid G divides each median in the ratio 2 : 1. xercise The centroid and the circumcenter of a triangle coincide if and only if the triangle is equilateral. 2.4.2 pollonius Theorem Theorem 2.5. Given triangle, let be the midpoint of. The length of the median is given by 2 + 2 = 2( 2 + 2 ).
142 The laws of sines and cosines Proof. pplying the law of cosines to triangles and, and noting that cos = cos, we have 2 = 2 + 2 2 cos; 2 = 2 + 2 2 cos, 2 = 2 + 2 + 2 cos. The result follows by adding the first and the third lines. If m a denotes the length of the median on the side, m 2 a = 1 4 (2b2 + 2c 2 a 2 ). xample. Suppose the medians and of triangle are perpendicular. This means that G 2 + G 2 = 2, where G is the centroid of the triangle. In terms of the lengths, we have 4 9 m2 b + 4 9 m2 c = a 2 ; 4(m 2 b +m2 c) = 9a 2 ; (2c 2 +2a 2 b 2 )+(2a 2 +2b 2 c 2 ) = 9a 2 ; b 2 + c 2 = 5a 2. This relation is enough to describe, given points and, the locus of for which the medians and of triangle are perpendicular. ere, however, is a very easy construction: rom b 2 + c 2 = 5a 2, we have m 2 a = 1 4 (2b2 + 2c 2 a 2 ) = 9 4 a2 ; m a = 3 a. The 2 locus of is the circle with center at the midpoint of, and radius 3. 2 2.4.3 The median triangle Given a triangle with sides a, b, c, the medians m a, m b, m c are the lengths of another triangle, which we call the median triangle of. In fact, if is extended to K such that = K, then (i) K is a parallelogram so that K =, (ii) K is also a parallelogram, and K =. The triangle K is a median triangle of ; its sides have lengths m a, m b, and m c. It has centroid. Note that the medians of this triangle are 3 4 m a, 3 4 m b, and 3 4 m c. In other words, the median triangle of the median triangle of is similar to. xample. (Triangles similar to their own median triangles) Now suppose we have a triangle which is similar to its own median triangle. Since the medians on longer sides of a triangle are shorter, if a b c, we must have m a : m b : m c = c : b : a.
2.4 The medians 143 G K m 2 b b 2 = m2 a c 2 = m2 c a 2 = m2 a + m2 c c 2 + a 2 = 1 4 (2a2 + 2b 2 c 2 ) + 1 4 (2a2 + 2b 2 c 2 ) = 3 c 2 + a 2 4. This means that m b = 3 b. Given and with midpoint, the locus of for which 2 is similar to its own median triangle is the circle with center the midpoint of, and radius the height of the equilateral triangle on : G Remarks. (1) The points,, G, are concyclic. (2) The bisectors of angles G and G intersect at right angles on G. xercise 1. The triangle with sides 7, 8, 9 has one median equal to a side. Which median and which side are these? 2 2. The lengths of the sides of a triangle are 136, 170, and 174. alculate the lengths of its medians. 3. Show that m 2 a + m2 b + m2 c = 3 4 (a2 + b 2 + c 2 ). 4. Triangle has sidelengths a = 17, b = 13, c = 7. Show that the medians are in the proportions of the sides. 2 nswer: m b = a = 7.
144 The laws of sines and cosines 7 13 17 5. (a) Show that the median m a can never be equal to the arithmetic mean of b and c. 3 (b) The median m a is the geometric mean of b and c if and only if a = 2 b c. c m a = bc b P 3 int: omplete the triangle to a parallelogram.