Showing two segments are congruent

Question

ABCD is a square. Draw the quarter circle centred at D with radius DA. Draw the semicircle of diameter AD. With point P on arc AC, $PQ \perp AB$, and $PD \cap \text{arc} AD = R$. Show that $PQ=PR$. enter image description here

Answer 1

Join $AR$ and $AP$. Let $E$ be the intersection between $AP$ and arc $AR$ such that $E\ne A$. Let $F$ denote the midpoint of $AD$.

Since $AB$ is tangent to $\odot D$ at $A$, we know that $\angle PAQ=\frac 12\angle ADP$. Also, $AB$ is tangent to $\odot F$ at $A$, we know that $\angle PAQ=\angle ADE$. Therefore, $\angle ADE=\angle EDR$. Since $\angle RAP=\angle EDR$, we conclude that $\angle RAP=\angle PAQ$. Since it is obvious that $\angle ARP=\angle AQP=90^\circ$ and $AP=AP$, $\triangle RAP$ is congruent to $\triangle QAP$, which yields that $RP=PQ$.