Competition problem — Geometry — Maths Olympiad Prep

Library / /103 of 129

Geometry Difficulty 6.0 National Olympiad Prove it Slovenia

Let $ABC$ be an acute triangle such that $|AB| > |BC| > |AC|$ . Let $D$ be a point different from $C$ on the segment $BC$ , such that $|AC| = |AD|$ . Let $H$ denote the orthocentre of the triangle $ABC$ , and let $A_1, B_1$ be the feet of the altitudes from $A$ and $B$ , respectively. The line $DH$ intersects the line $AC$ at $E$ and the line $A_1B_1$ at $F$ . Let $G$ be the intersection of the lines $AF$ and $BH$ . Show that the triangles $HBD$ and $HGE$ are similar.

Solution

The triangle $CAD$ is isosceles since $|AC| = |AD|$ . The line $AA_1$ is the altitude in this isosceles triangle, so $\angle HDA = \angle ACH$ .

In the quadrilateral $HA_1CB_1$ we have $\angle CA_1H = \frac{\pi}{2} = \angle CB_1H$ , so this quadrilateral is cyclic and $\angle B_1A_1H = \angle B_1CH$ . We have shown that $\angle FA_1A = \angle B_1A_1H = \angle B_1CH = \angle ACH = \angle HDA = \angle FDA$ , so $A$ , $D$ , $A_1$ and $F$ are concyclic. This implies that $\angle AFD = \angle AA_1D = \frac{\pi}{2}$ .

The segments $AB_1$ and $HF$ are the altitudes in the triangle $AHG$ and they meet at $E$ , so $E$ is the orthocentre of this triangle and $EG$ is perpendicular to $AH$ . Now, $AH$ is perpendicular to $BC$ , so $EG$ and $BC$ are parallel. Thus, $\angle EGH = \angle HBD$ and since $\angle GHE = \angle BHD$ we conclude that the triangles $HBD$ and $HGE$ are similar.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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