\begin{align} \angle ABD & = 90^\circ \phantom{000} (\text{Right } \angle \text{ in a semicircle)} \\ \\ \angle ABC & = 90^\circ \phantom{000} (\text{Right } \angle \text{ in a semicircle)} \\ \\ \angle DBC & = 90^\circ + 90^\circ \\ & = 180^\circ \\ \\ \therefore C, B & \text{ and } D \text{ lie on a straight line} \end{align}

\begin{align} \text{Let } & \angle APB = x^\circ \\ \\ \angle ABP & = \angle APB \phantom{000} \text{ (Isosceles triangle } APB \text{ since } AB = AP) \\ & = x^\circ \\ \\ \angle ABC & = 180^\circ - \angle ABP \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line)} \\ & = (180 - x)^\circ \\ \\ \angle CDA & = 180^\circ - \angle ABC \phantom{000} (\angle s \text{ in opposite segments)} \\ & = 180 - (180 - x) \\ & = 180 - 180 + x \\ & = x^\circ \\ & = \angle APB \\ & = \angle CPD \\ \\ \therefore \angle CDA & = \angle CPD \\ \\ \\ \text{Since } \angle CDA & = \angle CPD, \phantom{.} \triangle CDP \text{ is an isosceles triangle with } CD = CP \end{align}

(i)

\begin{align} \text{Let } & \angle AEF = x^\circ \\ \\ \angle ABF & = 180^\circ - \angle AEF \phantom{000} (\angle s \text{ in opposite segments)} \\ & = 180 - x \\ & = (180 - x)^\circ \\ \\ \angle ABC & = 180^\circ - \angle ABF \phantom{000} (\angle s \text{ on a straight line)} \\ & = 180 - (180 - x) \\ & = 180 - 180 + x \\ & = x^\circ \\ \\ \angle CDA & = 180^\circ - \angle ABC \phantom{000} (\angle s \text{ in opposite segments)} \\ & = 180 - x \\ & = (180 - x)^\circ \\ \\ \angle AEF + \angle CDA & = x + (180 - x) \\ & = 180^\circ \phantom{0} \text{ (Shown)} \end{align}

(ii)

\begin{align} \angle AEF + \angle CDA & = 180^\circ \phantom{000} (\text{From part i}) \\ \\ \therefore CD \phantom{.} & // \phantom{.} FE \phantom{000} (\text{Converse of interior angles}) \end{align}

\begin{align} \text{Since } PB \text{ and } PA \text{ are tangents to} & \text{ the circle at } B \text{ and } A \text{ respectively,} \\ PB & = PA \\ \\ \text{Since } PA \text{ and } PC \text{ are tangents to} & \text{ the circle at } A \text{ and } C \text{ respectively,} \\ PA & = PC \\ \\ \\ \therefore PB & = PA \\ PB & = PC \text{ (Shown)} \end{align}

(i)

\begin{align} \angle BAT & = \angle ACB \phantom{000} \text{ (Alternate segment theorem)} \\ \\ \angle ABC & = \angle ACB \phantom{000} \text{ (Isosceles triangle } ABC \text{ since } AB = AC) \\ & = \angle BAT \text{ (Shown)} \end{align}

(ii)

\begin{align} \angle ABC & = \angle BAT \phantom{000} (\text{From part i}) \\ \\ CB \phantom{.} & // \phantom{.} ST \phantom{000} (\text{Converse of alternate angles}) \end{align}

(i) Since OB, OC, OD and OA are the radii of the circle, triangles OBA, OCD and ODA are isosceles triangles.

\begin{align} \text{Let } & \angle OBC = x^\circ \\ \\ \angle OCB & = \angle OBC \phantom{000} \text{ (Isosceles triangle } OBC) \\ & = x^\circ \\ \\ \angle DOC & = \angle OCB \phantom{000} \text{ (Alternate } \angle s) \\ & = x^\circ \end{align}
(Consider triangles OAD and OCD) \begin{align} OD & \text{ is the common length [S]} \\ \\ \angle DOA & = \angle OBC \phantom{000} \text{ (Corresponding } \angle s) \\ & = x^\circ \\ & = \angle DOC \\ \\ \angle DOA & = \angle OBC \phantom{000} \text{ [A]} \\ \\ OA & = OC \phantom{000} \text{ (Radius of circle) [S]} \\ \\ \\ \triangle OAD & \text{ is congruent to } \triangle OCD \text{ (SAS)} \\ \\ \\ \therefore AD & = CD \phantom{000} (\triangle OAD \equiv \triangle OCD) \end{align}

(ii)

\begin{align} \angle COD & = x^\circ \phantom{000} (\text{From part i}) \\ \\ \angle ODC & = {180^\circ - x^\circ \over 2} \phantom{000} \text{ (Isosceles triangle } OCD) \\ & = \left( 90 - {1 \over 2}x\right)^\circ \\ \\ \\ \angle ODA & = \angle ODC \phantom{000} (\triangle OAD \equiv \triangle OCD) \\ & = \left( 90 - {1 \over 2}x\right)^\circ \\ \\ \angle CDE & = 180^\circ - \left( 90 - {1 \over 2}x\right)^\circ - \left( 90 - {1 \over 2}x\right)^\circ \phantom{000} (\angle \text{ sum of triangle)} \\ & = 180 - 90 + {1 \over 2}x - 90 + {1 \over 2}x \\ & = x^\circ \\ & = \angle COD \\ \\ \therefore \angle CDE & = \angle COD \phantom{0} \text{ (Shown)} \end{align}

(i)

(Consider triangles DOA and DOC) \begin{align} OD & \text{ is the common side [S]} \\ \\ \angle DOA & = \angle DOC \phantom{000} (D \text{ is the mid-point of arc } ADC) \text{ [A]} \\ \\ OA &= OC \phantom{000} (OA \text{ and } OC \text{ are the radii of the circle}) \text{ [S]} \\ \\ \therefore \triangle OAD & \text{ is congruent to } \triangle OCD \text{ (SAS)} \\ \\ \\ AD & = CD \phantom{000} (\triangle OAD \equiv \triangle OCD) \\ \\ \therefore \triangle DAC & \text{ is an isosceles triangle} \\ \\ \\ \text{Let } & \angle DAC = x^\circ \\ \\ \angle ADC & = 180^\circ - x^\circ - x^\circ \phantom{000} \text{ (Isosceles triangle } DAC) \\ & = (180 - 2x)^\circ \\ \\ \angle EDC & = 180^\circ - \angle ADC \phantom{000} (\angle s \text{ on a straight line)} \\ & = 180 - (180 - 2x) \\ & = 180 - 180 + 2x \\ & = 2(x) \\ & = 2\angle DAC \\ \\ \therefore \angle EDC & = 2\angle DAC \phantom{0} \text{ (Shown)} \end{align}

(ii)

\begin{align} \angle DAC & = x^\circ \phantom{000} (\text{From part i}) \\ \\ \angle CBD & = \angle DAC \phantom{000} (\angle s \text{ in the same segment)} \\ & = x^\circ \\ \\ \angle DCA & = \angle DAC \phantom{000} \text{ (Isosceles triangle)} \\ & = x^\circ \\ \\ \angle ABD & = \angle DCA \phantom{000} (\angle s \text{ in the same segment)} \\ & = x^\circ \\ & = \angle CBD \\ \\ \therefore \angle ABD & = \angle CBD \phantom{0} \text{ (Shown)} \end{align}

(i)

\begin{align} \angle APB & = 90^\circ \phantom{000} (\angle \text{ in semicircle)} \\ \\ \therefore \angle APB & = \angle PQA \phantom{000} \text{ [A]} \\ \\ \\ \angle BAQ & = 90^\circ \phantom{000} (\text{tangent} \perp \text{radius}) \\ \\ \angle BAQ + \angle PQA & = 90^\circ + 90^\circ \\ & = 180^\circ \\ \\ \therefore BA \phantom{.} & // \phantom{.} PQ \phantom{000} (\text{Converse of interior angles}) \\ \\ \angle BAP & = \angle APQ \phantom{0} \text{ (Alternate } \angle s, BA \phantom{.} // \phantom{.} PQ) \text{ [A]} \\ \\ \\ \therefore \triangle ABP & \text{ is similar to } \triangle PAQ \text{ (AA Similarity)} \end{align}

(ii)

\begin{align} \text{Since } \triangle ABP & \text{ is similar to } \triangle PAQ, \\ {AB \over PA} & = {AP \over PQ} \\ (AB)(PQ) & = (AP)(PA) \\ (AB)(QP) & = (AP)^2 \\ AP^2 & = AB \times QP \\ AP & = \pm \sqrt{AB \times QP} \\ & = \sqrt{AB \times QP} \phantom{0} \text{ or } -\sqrt{AB \times QP} \text{ (Reject)} \\ \\ \therefore AP & = \sqrt{AB \times QP} \phantom{0} \text{ (Shown)} \end{align}

\begin{align} \text{Let } & \angle ACB = x^\circ \\ \\ \angle BAF & = \angle ACB \phantom{000} \text{ (Alternate segment theorem)} \\ & = x^\circ \\ \\ \angle ABF & = \angle ACB \phantom{000} \text{ (Alternate segment theorem)} \\ & = x^\circ \\ \\ \angle AFB & = 180^\circ - \angle BAF - \angle ABF \phantom{000} (\angle \text{ sum of triangle)} \\ & = 180 - x - x \\ & = (180 - 2x)^\circ \\ \\ \angle BFD & = 180^\circ - \angle AFB \phantom{000} (\angle s \text{ on a straight line)} \\ & = 180 - (180 - 2x) \\ & = 180 - 180 + 2x \\ & = 2x \\ & = 2(x) \\ & = 2\angle ACB \\ \\ \therefore \angle BFD & = 2\angle ACB \phantom{0} \text{ (Shown)} \end{align}

(i)

\begin{align} \angle CAP & = \angle ABC \phantom{000} \text{ (Alternate segment theorem)} \\ & = \angle ABP \phantom{000} \text{ (Common } \angle s) \\ \\ \angle CAP & = \angle ABP \phantom{000} \text{ [A]} \\ \\ \angle APC & = \angle BPA \phantom{0} \text{ (Common } \angle s) \text{ [A]} \\ \\ \\ \therefore \triangle APC & \text{ is similar to } \triangle BPA \text{ (AA Similarity)} \end{align}

(ii)

\begin{align} \text{Since } \triangle APC & \text{ is similar to } \triangle BPA, \\ {AP \over BP} & = {PC \over PA} \\ {AP \over BP} & = {CP \over AP} \\ (AP)^2 & = (BP)(CP) \\ AP^2 & = BP \times CP \phantom{0} \text{ (Shown)} \end{align}

\begin{align} \text{Let } & \angle ANL = x^\circ \\ \\ \angle ALN & = \angle ANL \phantom{000} \text{ (Isosceles triangle, since } AL = AN) \\ & = x^\circ \\ \\ \angle BLP & = \angle ALN \phantom{000} \text{ (Vertically opposite } \angle s) \\ & = x^\circ \\ \\ \angle BPL & = \angle BPA \phantom{000} ( \text{Right } \angle \text{ in semicircle)} \\ & = 90^\circ \\ \\ \angle PBL & = 180^\circ - \angle BPL - \angle BLP \phantom{000} (\angle \text{ sum of triangle)} \\ & = 180 - 90 - x \\ & = (90 - x)^\circ \\ \\ \angle BAQ & = 90^\circ \phantom{000} \text{ (tangent} \perp \text{radius)} \\ \\ \angle ABL & = \angle ABN \\ & = 180^\circ - \angle BAQ - \angle ANL \phantom{000} (\angle \text{ sum of triangle)} \\ & = 180 - 90 - x \\ & = (90 - x)^\circ \\ & = \angle PBL \\ \\ \angle ABL & = \angle PBL \phantom{0} \text{ (Shown)} \end{align}

(i)

\begin{align} \angle PBL & = \angle ABN \phantom{000} (BN \text{ bisects } \angle ABP) \text{ [A]} \\ \\ \angle BAN & = 90^\circ \phantom{000} \text{ (tangent} \perp \text{radius)} \\ \\ \angle BPL & = \angle BPA \\ & = 90^\circ \phantom{000} (\text{Right }\angle \text{ in semicircle)} \\ & = \angle BAN \\ \\ \angle BPL & = \angle BAN \phantom{000} \text{ [A]} \\ \\ \\ \therefore \triangle BPL & \text{ is similar to } \triangle BAN \text{ (AA Similarity)} \end{align}

(ii)

\begin{align} \angle BLP & = \angle BNA \phantom{000} (\triangle BPL \text{ similar to } \triangle BAN) \\ & = \angle MNA \phantom{000} (\text{Common } \angle) \\ & = \angle ANM \\ \\ \angle BLP & = \angle ANM \phantom{000} \text{ [A]} \\ \\ \angle BMA & = 90^\circ \phantom{000} (\text{Right } \angle \text{ in semicircle)} \\ \\ \angle AMN & = 180^\circ - \angle BMA \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line)} \\ & = 180 - 90 \\ & = 90^\circ \\ & = \angle BPL \phantom{000} (\text{Right } \angle \text{ in semicircle)} \\ \\ \angle AMN & = \angle BPL \phantom{000} \text{ [A]} \\ \\ \\ \therefore \triangle BPL & \text{ is similar to } \triangle AMN \text{ (AA Similarity)} \end{align}

\begin{align} \text{Since } PS \text{ and } PU \text{ are tangents} & \text{ to the circle at } S \text{ and } U \text{ respectively,} \\ PS & = PU \\ \\ \text{Since } RS \text{ and } RT \text{ are tangents} & \text{ to the circle at } S \text{ and } T \text{ respectively,} \\ RS & = RT \\ \\ \\ \text{Since } QT \text{ and } QU \text{ are tangents} & \text{ to the circle at } T \text{ and } U \text{ respectively,} \\ QT & = QU \end{align}
\begin{align} 2PU & = PU + PU \\ & = PS + PU \phantom{000000000000000000} [PS = PU] \\ & = (PR + RS) + (PQ + QU) \\ & = PR + PQ + RS + QU \\ & = PR + PQ + RT + QT \phantom{00000000.} [RS = RT, QU = QT] \\ & = PR + PQ + (RT + QT) \\ & = PR + PQ + RQ \\ & = RP + PQ + QR \phantom{0} \text{ (Shown)} \end{align}

(i)

\begin{align} \angle BAC & = \angle CAP \phantom{000} \text{ (Common }\angle ) \text{ [A]} \\ \\ \angle ABC & = \angle CAX \phantom{000} \text{ (Alternate segment theorem)} \\ & = \angle ACP \phantom{000} \text{ (Alternate }\angle s) \\ \\ \angle ABC & = \angle ACP \phantom{000} \text{ [A]} \\ \\ \\ \therefore \triangle ABC & \text{ is similar to } \triangle ACP \text{ (AA Similarity)} \end{align}

(ii)

\begin{align} \text{Since } \triangle ABC & \text{ and } \triangle ACP \text{ are similar,} \\ {AB \over AC} & = {AC \over AP} \\ (AB)(AP) & = (AC)^2 \\ AP & = {AC^2 \over AB} \\ \\ \therefore PA & = {AC^2 \over AB} \end{align}

\begin{align} \text{Let } & \angle CAX = x^\circ \\ \\ \angle ACP & = \angle CAX \phantom{000} \text{ (Alternate }\angle s) \\ & = x^\circ \\ \\ \angle PCB & = \angle ACP \phantom{000} (PC \text{ bisects } \angle BCA) \\ & = x^\circ \\ \\ \angle PBC & = \angle CAX \phantom{000} \text{ (Alternate segment theorem)} \\ & = x^\circ \\ & = \angle PCB \\ \\ \\ \text{Since } \angle PCB = \angle PBC, \phantom{.} & \triangle PBC \text{ is an isosceles triangle} \\ \\ \therefore PB & = PC \text{ (Shown)} \end{align}

(i)

\begin{align} \angle EBC & = \angle ADC \phantom{000} (\angle s \text{ in the same segment) [A]} \\ \\ \angle BEC & = 90^\circ \\ & = \angle DAC \phantom{000} (\angle \text{ in semicircle)} \\ \\ \angle BEC & = \angle DAC \phantom{000} \text{ [A]} \\ \\ \\ \therefore \triangle EBC & \text{ is similar to } \triangle ADC \text{ (AA Similarity)} \end{align}

(ii)

\begin{align} \text{Since } \triangle EBC & \text{ is similar to } \triangle ADC, \\ \\ {BC \over DC} & = {EC \over AC} \\ \\ {BC \over 2r} & = {EC \over AC} \\ \\ {(BC)(AC) \over 2r} & = EC \\ \\ \therefore EC & = {BC \times AC \over 2r} \\ \\ \text{Area of } \triangle ABC & = {1 \over 2} \times Base \times Height \\ & = {1 \over 2} \times AB \times EC \\ & = {1 \over 2} \times AB \times {BC \times AC \over 2r} \\ & = {AB \over 2} \times {BC \times AC \over 2r} \\ & = {AB \times BC \times AC \over 4r} \\ & = {AB \times BC \times CA \over 4r} \phantom{0} \text{ (Shown)} \end{align}

(i)

\begin{align} PQ^2 & = PC \times PA \phantom{000000} [\text{Tangent-secant theorem}] \\ \\ \text{Since } PQ^2 = & \phantom{.} AP \times BR, \\ AP \times BR & = PC \times PA \\ BR & = PC \\ \\ \implies AR & = AP \\ \\ \\ QR^2 & = BR \times AR \phantom{000000} [\text{Tangent-secant theorem}] \\ \\ \angle AQR & = 90^\circ \phantom{00000000000} [\text{Tangent perpendicular to radius}] \\ \\ \text{By Pyth} & \text{agoras theorem,} \\ AQ^2 & = AR^2 - QR^2 \\ & = AR^2 - BR \times AR \\ & = AR(AR - BR) \\ & = AR(AB) \\ & = AP(AB) \\ & = AP \times AB \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} \angle AQP & = \angle AQR = 90^\circ \phantom{000} [R] \\ \\ AP & = AR \phantom{00000000000} [H] \\ \\ AQ \text{ is a } & \text{common side } [S] \\ \\ \text{Triangles } AQP & \text{ and } AQR \text{ are congruent (RHS)} \\ \\ \\ \angle ABQ & = 90^\circ \phantom{0} (\text{Right angle in semi-circle}) \\ \angle ACQ & = 90^\circ \phantom{0} (\text{Right angle in semi-circle}) \\ \\ \angle ABQ & = \angle ACQ \phantom{00000} [R] \\ \\ AQ \text{ is a } & \text{common side } \phantom{00} [H] \\ \\ \text{From (i), } & AP = AR \text{ and } BR = CP, \\ AB & = AR - BR \\ & = AP - CP \\ & = AC \phantom{00000000} [S] \\ \\ \text{Triangles } ABQ & \text{ and } ACQ \text{ are congruent (RHS)} \end{align}

(iii)

\begin{align} \angle ABQ & = \angle AQR \phantom{00} [A] \\ \\ \angle BAQ & = \angle QAR \phantom{0} \text{(Common angle) } [A] \\ \\ \text{Triangles } & ABQ \text{ and } AQR \text{ are similar (AA)} \\ \\ {AB \over AQ} & = {AQ \over AR} \\ AB \times AR & = AQ^2 \\ AB \times AR & = AP \times AB \phantom{0000} [\text{part (i)}] \\ \\ \text{Since triangles } & ABQ \text{ and } ACQ \text{ are congruent, } AB = AC \\ \\ \therefore AC \times AR & = AP \times AB \end{align}

\begin{align} & \text{Note: The tangents are perpendicular } \text{to the radii} \\ \\ \text{Area of triangle } ABC & = \text{Area of triangle } ABO + \text{Area of triangle } ACO + \text{Area of triangle } BCO \\ & = {1 \over 2} \times AB \times r + {1 \over 2} \times AC \times r + {1 \over 2} \times BC \times r \\ & = {1 \over 2} r (AB + AC + BC) \\ & = {1 \over 2} r (2s) \\ & = rs \phantom{00} \text{ (Shown)} \end{align}