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Problem 1165

AIME late

Geometry Difficulty 5.1 Prove it Irish Mathematical Olympiad · Ireland

In triangle ABC the perpendicular projection $BD$ of $AB$ on $BC$ has the same length as the perpendicular projection $CH$ of $BC$ on $CA$ . Prove that the altitude from $A$ , the median from $B$ and the bisector of $\angle ACB$ are concurrent.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $E$ be the midpoint of $CA$ and $F$ the intersection point of $AB$ and the bisector of $\angle ACB$ . We will use (the converse of) Ceva's Theorem to prove that $AD$ , $BE$ , $CF$ are concurrent.

We first note that $A$ , $B$ , $D$ , $H$ are concyclic because of the right angles at $D$ and $H$ . The Intersecting Secants Theorem then implies
$\begin{aligned} |CD| \cdot |CB| &= |CH| \cdot |CA|, \text{ i.e.} \\ \frac{|CH|}{|DC|} &= \frac{|CB|}{|CA|} \end{aligned}$
Using $|CH| = |BD|$ and that the angle bisector $CF$ divides $AB$ in the ratio of the adjacent sides, we obtain
$\frac{|BD|}{|DC|} = \frac{|FB|}{|AF|}, \text{ or } \frac{|BD|}{|DC|} \cdot \frac{|AF|}{|FB|} = 1.$
Because $|CE| = |EA|$ , we finally obtain
$\frac{|BD|}{|DC|} \cdot \frac{|CE|}{|EA|} \cdot \frac{|AF|}{|FB|} = 1$
which implies that $AD$ , $BE$ , $CF$ are concurrent.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.