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Berkeley Math Circle Monthly Contest 5

Geometry Difficulty 4.5 AIME Prove it United States

Problem:

In triangle $A B C$ , the bisector of $\angle B$ meets the circumcircle of $\triangle A B C$ at $D$ . Prove that
$B D^{2}>B A \cdot B C$

Solution

Solution:

In the diagram, $\angle A B D = \angle D B C$ (angle bisector) and $\angle B A C = \angle B D C$ (both intercept arc $B C$ ), so $\triangle B A E \sim \triangle B D C$ . We get
$\begin{aligned} \frac{B A}{B E} & = \frac{B D}{B C} \\ B D \cdot B E & = B A \cdot B C, \end{aligned}$
from which the desired inequality follows since $B E < B D$ .

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