1969 Canadian MO Problems/Problem 5

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 •	Followed by Problem 6

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1969 Canadian MO Problems/Problem 5

Problem

Solution