Library / Iranian Mathematical Olympiad /7 of 270

Iranian Mathematical Olympiad

Geometry Difficulty 5.1 AIME, harder Prove it Iran

In quadrilateral $ABCD$ , $AC$ is the bisector of angle $A$ and $\angle ADC = \angle ACB$ . $X$ and $Y$ are the feet of the altitudes from $A$ to $BC$ and $CD$ , respectively. Prove that the orthocenter of triangle $AXY$ is located on line $BD$ (A triangle's orthocenter is the intersection point of its altitudes).

Solution

Let $E$ be the foot of the perpendicular line from $Y$ to $AX$ and $P$ be the intersection point of $YE$ and $BD$ . It is sufficient to prove that $XP \perp AY$ , or equivalently, $XP \parallel CD$ . Note that
$YE \parallel CB \Rightarrow \frac{DY}{YC} = \frac{DP}{PB}. \quad (1)$
On the other hand, triangles $ADC$ and $ACB$ are similar, so
$\frac{DY}{YC} = \frac{CX}{XB}. \quad (2)$

(1) and (2) show that $\frac{DP}{PB} = \frac{CX}{XB}$ , and this implies $XP \parallel CD$ , as desired.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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