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 2 of 5: Identify

D'

with

A

Previous step Next step

D'=A

Detailed analysis

Quadrilateral $APDQ$ is cyclic because $\angle APD$ and $\angle AQD$ are right angles, so $PQ$ crosses $AD$ between $A$ and $D$ . Since $PQ\perp AD$ , reflecting $D$ across $PQ$ puts $D'$ on the ray from $D$ through $A$ . The line $AD$ meets the circumcircle at $A$ and at a second point beyond $D$ (because $D$ lies on the chord $BC$ ), so the only circumcircle point on the ray $DA$ is $A$ ; hence $D'=A$ .