Problem 1

We say that a triangle

ABC

is great if the following holds: for any point

D

on side

BC

, if

P

and

Q

are the feet of the perpendiculars from

D

to the lines

AB

and

AC

respectively, then the reflection of

D

in the line

PQ

lies on the circumcircle of triangle

ABC

. Prove that triangle

ABC

is great if and only if

\angle A=90^\circ

and

AB=AC

Step 5 of 5: Sufficiency: right isosceles triangles are great

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\angle BAC=90^\circ,\ AB=AC \implies \angle BD'C=\angle PDQ=90^\circ

Detailed analysis

Conversely, suppose $\angle BAC=90^\circ$ and $AB=AC$ , and let $D$ be any point of $BC$ with $P$ , $Q$ its projections onto $AB$ , $AC$ and $D'$ the reflection of $D$ in $PQ$ . Since $D'P=DP=BP$ and $D'Q=DQ=CQ$ , and $APDQD'$ lies on a circle with diameter $PQ$ , one gets $\angle BPD'=\angle CQD'$ , so triangles $D'PB$ and $D'QC$ are similar; this yields $\angle PD'Q=\angle BD'C$ and, since triangles $DPQ$ and $D'PQ$ are congruent, $\angle BD'C=\angle PDQ=90^\circ$ . Hence $D'$ lies on the circle with diameter $BC$ , which is exactly the circumcircle of the right isosceles triangle $ABC$ , proving $ABC$ is great.