Library / Taiwan IMO Selection Camp /2 of 359

Taiwan IMO Selection Camp

Geometry Difficulty 4.4 AIME Prove it Taiwan

Let $O$ be the circumcenter of acute triangle $\triangle ABC$ , and let line $AO$ meet $BC$ at a point $D$ . Take points $E, F$ on sides $AB$ and $AC$ respectively such that $ED = BD$ and $DF = CD$ . Prove that $EF \parallel BC$ .

Solution

First, draw line $AO$ to meet circle $O$ at $P$ , then $\angle ABP = 90^\circ = \angle ACP$ .

Next, draw the perpendiculars from point $D$ to sides $BE$ and $CF$ , meeting $BE$ and $CF$ at points $G$ and $H$ respectively. Since $\triangle AGD \sim \triangle ABP$ and $\triangle AHD \sim \triangle ACP$ , we have $AG/AB = AD/AP = AH/AC$ . Therefore $\triangle AGH \sim \triangle ABC$ , hence $GH \parallel BC$ .

Finally, since $ED = BD$ and $DF = CD$ , we have $BG = GE$ and $CH = HF$ . Connect $CE$ meeting $GH$ at point $M$ . Then since $GH \parallel BC$ , we know $GM \parallel BC$ , hence $EM = MC$ . Also $CH = HF$ , therefore $EF \parallel HF$ , hence $EF \parallel BC$ .

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 translated into English from zh; metadata (topic, difficulty) added by this project.