Question

In the figure, PBX and QBY are segments and angle PAB = angle QAB.
Prove that PQXY is cyclic.​

Answer 1

Since $\angle QAB \cong \angle PAB$ and by inscribed angle theorem,

$\angle PAB \cong \angle PYB;\\

\angle QAB \cong \angle QXB$

By transitivity,

$\implies \angle QXB \cong \angle PYB\\

\mathrm {or}\: \angle PXQ \cong \angle QYP\\

\implies \text{PQXY is cyclic}$

This follows from the converse of the theorem that angles subtended on a circle by a chord of fixed length are equal. Here, chord $PQ$ of the circle passing through $PQXY$ subtends equal angles at points $X$ and $Y$ on the circle.