Competition problem — Geometry — Maths Olympiad Prep

Library / /22 of 53

Geometry Difficulty 6.0 AIME, harder Prove it China

Suppose that the convex quadrilateral $ABCD$ satisfies $AB = BC$ , $AD = DC$ . $E$ is a point on $AB$ , and $F$ on $AD$ , such that $B$ , $E$ , $F$ , $D$ are concyclic. Draw $\triangle DPE$ directly similar to $\triangle ADC$ , and $\triangle BQF$ directly similar to $\triangle ABC$ . Prove that $A$ , $P$ , $Q$ are collinear. (Posed by Ye Zhonghao)

Denote by $O$ the center of the circle that passes through $B$ , $E$ , $F$ , $D$ . Draw lines $OB$ , $OF$ , $BD$ .

Solution

In $\triangle BDF$ , $O$ is the circumcenter, so $\angle BOF = 2\angle BDA$ ; And $\triangle ABD \sim \triangle CBD$ , so $\angle CDA = 2\angle BDA$ . Hence, $\angle BOF = \angle CDA = \angle EPD$ , which implies that the isosceles triangles
$\triangle BOF \sim \triangle EPD. \qquad \textcircled{1}$
On the other hand, the concyclicity of $B$ , $E$ , $F$ , $D$ implies that
$\triangle ABF \sim \triangle ADE. \qquad \textcircled{2}$
Combining ① and ②, we know that the quadrilateral $ABOF \sim ADPE$ , so
$\angle BAO = \angle DAP. \qquad \textcircled{3}$
The same argument gives
$\angle BAO = \angle DAQ. \qquad \textcircled{4}$
③ and ④ imply that $A$ , $P$ , $Q$ are collinear.

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 and solution reproduced as published; topic and difficulty added by this site.