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 1 of 5: Choose

D

on the bisector from

A

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PQ \perp AD

Detailed analysis

Assume $ABC$ is great. Let $D$ be the point where the bisector of $\angle BAC$ meets $BC$ , and let $P$ , $Q$ be the feet of the perpendiculars from $D$ to lines $AB$ , $AC$ . Since $AD$ bisects $\angle A$ , points $P$ and $Q$ lie on rays $AB$ , $AC$ and are reflections of each other across line $AD$ ; hence line $PQ$ is perpendicular to $AD$ . Let $D'$ be the reflection of $D$ in line $PQ$ ; by the great condition, $D'$ lies on the circumcircle of $ABC$ .