\begin{align*} \angle ADB & = \angle ACB = 90^\circ \phantom{0} (\text{Right angle in semi-circle}) \\ \\ \angle PDB & = \angle QDB = 90^\circ \phantom{0} [R] \\ \\ BP & = BQ \phantom{0} \text{(Given)} [H] \\ \\ BD & \text{ is a common side} [S] \\ \\ \text{Triangles } & BDP \text{ and } BDQ \text{ are similar (RHS congruency test)} \\ \\ \implies \angle DBP & = \angle DBQ \\ \\ \therefore DB & \text{ bisects angle } PBQ \end{align*}

\begin{align*} \angle AOP & = \angle ADB = 90^\circ \phantom{0} [A] \\ \\ \angle OAP & = \angle DAB \phantom{0} \text{(Common angle})[A] \\ \\ \therefore \text{Triangles } & AOP \text{ and } ADB \text{ are similar (AA similarity test)} \end{align*}

\begin{align*} \therefore \text{Triangles } & AOP \text{ and } ADB \text{ are similar} \\ \\ \implies {AO \over AD} & = {AP \over AB} \\ AO \times AB & = AP \times AD \\ AO \times 2 AO & = AP \times AD \\ 2 AO^2 & = AP \times AD \\ AO^2 & = {1 \over 2} \times AP \times AD \phantom{0} \text{ (Shown)} \end{align*}

\begin{align*} \text{Let } \angle BAP & = \theta \\ \\ \angle BCA & = \theta \phantom{0} (\text{Alternate segment theorem}) \\ \\ \angle DAC & = \theta \phantom{0} (\text{Alternate angles, } DA \phantom{.} // \phantom{.} CB) \\ \\ \angle DBC & = \theta \phantom{0} (\text{Angles in the same segment}) \\ \\ \angle DCA & = \theta \phantom{0} (\text{Isosceles triangle}) \\ \\ \angle DBA & = \theta \phantom{0} (\text{Angles in the same segment}) \\ \\ \\ \angle ABP & = 180^\circ - \theta - \theta \phantom{0} (\text{Adjacent angles on a straight line}) \\ & = 180^\circ - 2\theta \\ \\ \angle BPA & = 180^\circ - (180^\circ - 2\theta) - \theta \phantom{0} (\text{Angle sum of triangle}) \\ & = 180^\circ - 180^\circ + 2 \theta - \theta \\ & = \theta \\ & = \angle BAP \\ \\ \therefore \text{Triangle } & PAB \text{ is isosceles} \end{align*}

\begin{align*} \text{From (i), } \angle PAB & = \angle DBA = \theta \\ \\ \text{By the converse of alternate} & \text{ angles, lines } AP \text{ and } DB \text{ are parallel} \end{align*}