hapter 3 The angle bisectors 3.1 The angle bisector theorem Theorem 3.1 (ngle bisector theorem). The bisectors of an angle of a triangle divide its opposite side in the ratio of the remaining sides. If and respectively the internal and external bisectors of angle, then : = c : b and : = c : b. c b Proof. onstruct lines through parallel to the bisectors and to intersect the line at and. (1) Note that = = =. This means =. learly, : = : = : = c : b. () Similarly, =, and : = : = : = c : b.
146 The angle bisectors Exercise 1. Given a segment, erect a square on it, and an adjacent one with base. If D is the vertex above, construct the bisector of angle D to intersect at P. alculate the ratio P : P. D P. square is inscribed in a right triangle with sides a and b. Show that each side of the square has length l = ab a + b. a l l b 3. In the diagram, D, D, Q, and P are squares. Show that, P,, Q are concyclic. Q D D P
3.1 The angle bisector theorem 147 4. Given a right triangle with a right angle at and α β, let P be the intersection of the median on and the bisector of angle. Show that lines through P perpendicular to the sides of the triangle cut out triangles of equal areas. D F P 5. Let be a triangle with =. The internal bisector of angle meets at D. Let M and N be the midpoints of and D respectively. Suppose that, M, D, N are concyclic. Prove that = 7. 6. In triangle, α = 10. is the bisector of angle. Show that 1 t = 1 b + 1 c. c t b 7. In the diagram below,,, and D are equilateral triangles. Suppose = 10. Show that 1 b = 1 a + 1 c. 1 a b c D 1 Hint: Extend to intersect at T. Show that T = a.
148 The angle bisectors 3. Steiner-Lehmus theorem Theorem 3.. triangle is isosceles if it has two equal angle bisectors. F E Proof. Suppose the bisectors E = F, but triangle not isosceles. We may assume <. onstruct parallels to through E and F to intersect and at and respectively. (1) In the isosceles triangles E and F with equal bases E and F, E < F = E < F. () = F = < = E. Therefore, F E = + 1 < E + 1 =, and > E. This clearly implies F < E, E E contradicting (1) above. Example. The bisectors of angles and of triangle intersect the median D at E and F respectively. Suppose E = F. Show that triangle is isosceles. α α E F D Solution. We first show that ED = FD. pplying the law of sines to triangles ED and FD, we have D sin ED = E sin DE = F sin DF = D sin FD. Since D = D, we conclude that sin ED = sin FD. The two angles are either equal or supplementary. Since the sum of these two angles is + + < 180, they must be equal.
3. Steiner-Lehmus theorem 149 If triangle is not isosceles, say b > c, then the bisector of angle intersects at, then < D. This means that α > α. It follows that a contradiction. Exercise ED = α + > α + = FD, 1. is a triangle with both external angle bisectors t b and t c equal to a. alculate the angles of the triangle. t b a t c. The bisector t a and the external bisector t b of triangle satisfy t a = t b = c. alculate the angles of the triangle. t b t a c
150 The angle bisectors 3..1 The lengths of the bisectors Proposition 3.1. (a) The lengths of the internal and external bisectors of angle are respectively t a = bc b + c cos α and t a = bc b c sin α. c t a b t a Proof. Let and be the bisectors of angle. (1) onsider the area of triangle as the sum of those of triangles and. We have 1 t a(b + c) sin α = 1 bc sin α. From this, t a = bc b + c sin α sin α = bc b + c cos α. () onsider the area of triangle as the difference between those of and. Remarks. (1) bc is the harmonic mean of b and c. It can be constructed as follows. If the b+c perpendicular to at intersects and at and, then = = bc. b+c c t a b
3. Steiner-Lehmus theorem 151 () pplying Stewart s Theorem with λ = c and µ = ±b, we also obtain the following expressions for the lengths of the angle bisectors: ( ( ) ) a t a = bc 1, b + c ( ( ) t a a = bc 1). b c lternative proof of Steiner-Lehmus theorem We show that if a < b, then t a > t b. Note that from a < b we conclude (i) α < β and cos α > cos β ; b (ii) bc > ac, b(c + a) > a(b + c); > a ; bc > ca. b+c c+a b+c c+a From (i) and (ii), we have t a = bc b + c cos α > ca c + a cos β = t b. The same reasoning shows that a > b t a < t b. It follows that if t a = t b, then a = b. Exercise 1. The lengths of the sides of a triangle are 84, 15, 169. alculate the lengths of its internal bisectors.. is a right triangle in which the bisector of the right angle, and the median to the hypotenuse have lengths 4 and 35 respectively. alculate the sidelengths of the triangle. 3. (a) In triangle, a = 5, b = 8, c = 7. Show that t a : t b = b : a. 5 8 t a t b 7 (b) Suppose t a : t b = b : a. Show that the triangle is either isosceles, or γ = 60. nswers: 975 7, 608 53, 1600 09.
15 The angle bisectors 4. (a) In triangle, a = 7, b = 5 and c = 3. Let,, be the angle bisectors. Show that is a right triangle. (b) If,, are the angle bisectors of triangle, show that + = a bc(b + bc + c a ) (b + c) (c + a)(a + b). I (c) Given a segment, construct the locus of for which = 90. 5. Find a right triangle for which the bisector of an acute angle is the geometric mean of the two segments it divides on the opposite side. a t c E 6. Find an isosceles triangle for which the bisector of a base angle is the geometric mean of the two segments it divides on the opposite side. 3 3 nswer: (a, b, c) = (1, 1 +, 1 + ).
3. Steiner-Lehmus theorem 153 7. In, α = 60, and β < γ. The bisector of intersects at. If is a mean proportional between and, find β. 8. The bisector of angle of triangle intersects at. Show that is the geometric mean of and if and only if b + c = a. 9. Let t a, t b, t c be the lengths of the bisectors of a triangle, and T a, T b, T c these angle bisectors extended until they are chords of the circumcircle. Prove that abc = t a t b t c T a T b T c. I
154 The angle bisectors Excursus: Triangles with two equal external angle bisectors Theorem 3.3. If t b = t c, then sin α = sin β sin γ. t b a t c Proof. ssume t b = t c. In triangles and, we have = α γ and = β α. y the law of sines, Thus, sin β sin γ sin α γ sin β sin α γ sin β cos β sin α γ ( sin α + β γ sin β + γ α ) = t b a = t c a = sin β sin β α = sinγ sin β α = sin γ cos γ sin β α = sin γ. ( sin β + γ α sin γ + β α sin β (cos γ cos α) = sin γ (cos α cos β) sin β ( sin α γ ) sin = sin γ ( sin β α ) sin sin β sin γ ( sin β + sin γ ) = sin α ( sin β + sin γ ). ) The result follows from canceling the nonzero common divsior sin β + sin γ. Remark. Equivalently, if t b = t c, then ( ) s a = s b s c. a b c
3.3 The circle of pollonius 155 3.3 The circle of pollonius Theorem 3.4. and are two fixed points. For a given positive number k 1, 4 the locus of points P satisfying P : P = k : 1 is the circle with diameter, where and are points on the line such that : = k : 1 and : = k : 1. P O Proof. Since k 1, points and can be found on the line satisfying the above conditions. onsider a point P not on the line with P : P = k : 1. Note that P and P are respectively the internal and external bisectors of angle P. This means that angle P is a right angle, and P lies on the circle with as diameter. onversely, let P be a point on this circle. We show that P : P = k : 1. Let be a point on the line such that P bisects angle P. Since P and P are perpendicular to each other, the line P is the external bisector of angle P, and = = =. On the other hand, = = =. omparison of the two expressions shows that coincides with, and P is the bisector of angle P. It follows that P = = k. P 4 If k = 1, the locus is clearly the perpendicular bisector of the segment.
156 The angle bisectors Exercise 1. If = d, and k 1, the radius of the pollonius circle is k k 1 d.. Suppose is a triangle with, and let D, E,, be points on the line defined as follows: D is the midpoint of, E is the foot of the perpendicular from to,, bisect angle. Prove that = DE. E D 3. Given two disjoint circles () and (), find the locus of the point P such that the angle between the pair of tangents from P to () and that between the pair of tangents from P to () are equal. P