MathLabs

Problem 1

We say that a triangle ABCABC is great if the following holds: for any point DD on side BCBC, if PP and QQ are the feet of the perpendiculars from DD to the lines ABAB and ACAC respectively, then the reflection of DD in the line PQPQ lies on the circumcircle of triangle ABCABC. Prove that triangle ABCABC is great if and only if ∠A=90∘\angle A=90^\circ and AB=ACAB=AC.
Step 3 of 5: Angle chase forces ∠A=90∘\angle A=90^\circ
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∠PD′Q=∠PDQ=180∘−∠BAC,∠PD′Q=∠BAC  ⟹  ∠BAC=90∘\angle PD'Q=\angle PDQ=180^\circ-\angle BAC,\quad \angle PD'Q=\angle BAC \implies \angle BAC=90^\circ
Detailed analysis

Reflecting across PQPQ preserves angles, so ∠PD′Q=∠PDQ\angle PD'Q=\angle PDQ, and since APDQAPDQ is cyclic with right angles at PP and QQ, ∠PDQ=180∘−∠BAC\angle PDQ=180^\circ-\angle BAC. But D′=AD'=A gives ∠PD′Q=∠PAQ=∠BAC\angle PD'Q=\angle PAQ=\angle BAC. Combining the two expressions for ∠PD′Q\angle PD'Q yields ∠BAC=90∘\angle BAC=90^\circ.