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1969 Canadian MO Problems/Problem 5

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Problem

Let ABC be a triangle with sides of length a, b and c. Let the bisector of the \angle C cut AB at D. Prove that the length of CD is \frac{2ab\cos \frac{C}{2}}{a+b}.


Solution

Let CD=d. Note that [\triangle ABC]=[\triangle ACD]+[\triangle BCD]. This can be rewritten as \frac12 a b \sin C=\frac12 a d \sin \frac C2 +\frac12 b d \sin \frac C2 .

Because \sin C=2\sin \frac C2 \cos \frac C2, the expression can be written as 2ab\cos \frac C2=d(a+b). Dividing by a+b, CD=\frac{2ab\cos \frac C2}{a+b}, as desired.

1969 Canadian MO (Problems)
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 Followed by
Problem 6


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