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 Followed by
Problem 6


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