(i)

\begin{align} AH & = GB \phantom{000} \text{ (Given)} \\ AG + GH & = BA + AG \\ GH & = BA + AG - AG \\ & = BA \\ \\ GH & = AB \phantom{000} \text{ [S]} \\ \\ \angle GHI & = \angle ABC \phantom{000} \text{ (Alternate }\angle s) \text{ [A]} \\ \\ HI & = BC \phantom{000} \text{ (Given) [S]} \\ \\ \\ \therefore \triangle GHI & \text{ is congruent to } \triangle ABC \text{ (SAS)} \end{align}

(ii)

\begin{align} \text{Let } & \angle BAC = x^\circ \\ \\ \angle HGI & = \angle BAC \phantom{000} (\triangle GHI \equiv \triangle ABC) \\ & = x^\circ \\ \\ \angle GAC & = 180^\circ - \angle BAC \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line)} \\ & = 180^\circ - x^\circ \\ & = 180^\circ - \angle HGI \\ & = \angle AGI \\ \\ \angle GAC & = \angle AGI \\ \\ \therefore CA \phantom{.} & // \phantom{.} GI \phantom{000} (\text{Converse of alternate } \angle s) \end{align}

(i)

\begin{align} \angle APQ & = \angle ABC \phantom{000} \text{ (Given) [A]} \\ \\ \angle PAQ & = \angle BAC \phantom{000} \text{ (Common }\angle) \text{ [A]} \\ \\ \\ \therefore \triangle APQ & \text{ is similar to } \triangle ABC \text{ (AA Similarity)} \end{align}

(ii)

\begin{align} \text{Since } \triangle APQ & \text{ is similar to } \triangle ABC, \\ {AP \over AB} & = {AQ \over AC} \\ {(AP)(AC) \over AB} & = AQ \\ {\left({1 \over 2}AC\right)(AC) \over AB} & = AQ \phantom{000000} [P \text{ is the mid-point of } AC] \\ {{1 \over 2}AC^2 \over AB} & = AQ \\ \\ AQ & = {{1 \over 2}AC^2 \over AB} \\ AQ & = {{1 \over 2}AC^2 \over AB} \times {2 \over 2} \\ AQ & = {AC^2 \over 2AB} \phantom{0} \text{ (Shown)} \end{align}

(i)

\begin{align} \angle AEF & = \angle DEG \phantom{000} \text{ (Vertically opposite } \angle s) \text{ [A]} \\ \\ \angle EAF & = \angle EDG \phantom{000} \text{ (Alternate } \angle s) \text{ [A]} \\ \\ \\ \therefore \triangle AEF & \text{ is similar to } \triangle DEG \text{ (AA Similarity)} \end{align}

(ii)

\begin{align} \text{Since } \triangle AEF & \text{ is similar to } \triangle DEG, \\ {AE \over DE} & = {AF \over DG} \\ {AE \over 2AE} & = {AF \over DG} \phantom{00000} \left[ \text{Since } AE = {1 \over 3}AD, \phantom{.} DE = 2AE\right] \\ {1 \over 2} & = {AF \over DG} \\ \\ DG & = 2AF \end{align}
\begin{align} \angle BCF & = \angle DCG \phantom{000} (\text{Common } \angle) \text{ [A]} \\ \\ \angle BFC & = \angle DGC \phantom{000} (\text{Corresponding } \angle s, \phantom{.} BF \phantom{.} // \phantom{.} DG) \text{ [A]} \\ \\ \\ \triangle BCF & \text{ is similar to } \triangle DCG \text{ (AA Similarity)} \\ \\ \\ \text{Since } \triangle BCF & \text{ is similar to } \triangle DCG, \\ {BF \over DG} & = {BC \over DC} \\ {BF \over DG} & = {2DC \over DC} \phantom{000} (D \text{ is the mid-point of } BC) \\ {BF \over DG} & = 2 \\ BF & = 2DG \\ BF & = 2(2AF) \phantom{00000} [DG = 2AF] \\ & = 4AF \phantom{0} \text{ (Shown)} \end{align}

(iii)

\begin{align} BF & = 4AF \phantom{000} (\text{From part ii}) \\ & = 4(1) \\ & = 4 \text{ cm} \\ \\ AB & = AF + BF \\ & = 1 + 4 \\ & = 5 \text{ cm} \end{align}

(i)

\begin{align} PD & = QB \phantom{000} \text{ (Given) [S]} \\ \\ \angle PDC & = \angle QBA \phantom{000} \text{(Opposite angles of parallelogram are equal) [A]} \\ \\ DC & = BA \phantom{000} \text{(Opposite sides of parallelogram are equal) [S]} \\ \\ \\ \triangle PDC & \text{ is congruent to } \triangle QBA \text{ (SAS)} \end{align} $$ \therefore \angle DPC = \angle BQA \phantom{00} (\triangle PDC \equiv \triangle QBA) $$

(ii)

\begin{align} \angle CPA & = 180^\circ - \angle DPC \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line)} \\ & = 180^\circ - \angle BQA \phantom{000} (\text{From part i}) \\ & = \angle AQC \\ \\ \text{Let } & \angle DAB = x^\circ \\ \\ \angle DCB & = \angle DAB \phantom{000} \text{(Opposite angles of parallelogram are equal)} \\ & = x^\circ \\ \\ \angle PCD & = \angle QAB \phantom{000} (\triangle PDC \equiv \triangle QBA) \\ \\ \angle PCQ & = \angle DCB - \angle PCD \\ & = x^\circ - \angle PCD \\ & = x^\circ - \angle QAB \\ & = \angle DAB - \angle QAB \\ & = \angle PAQ \\ \\ \therefore \angle PCQ & = \angle PAQ \end{align} \begin{align} \text{Since } \angle CPA = \angle AQC \text{ and } \angle PCQ = \angle PAQ, \phantom{.} AQCP \text{ is a parallelogram} \end{align}

(i)

\begin{align} \text{If } \angle SRY & = \angle RYP = 90^\circ, \phantom{.} RX = XY \\ \\ \therefore XU \phantom{.} & // \phantom{.} YQ \phantom{.} (\text{Midpoint theorem}) \\ \\ \implies TU \phantom{.} & // \phantom{.} PQ \phantom{.} // \phantom{.} SR \\ \\ \\ \text{Since } RU = UQ \text{ and } VU \phantom{.} & // \phantom{.} PQ, \text{ by the converse of } \text{midpoint theorem,} \\ RV & = VP \end{align}
\begin{align} TU & = TV + VU \\ & = \left({1 \over 2}SR\right) + VU \phantom{000} \text{(Mid-point theorem)} \\ & = {1 \over 2}SR + VU \\ & = {1 \over 2}(PY) + VU \\ & = {1 \over 2}\left({1 \over 2}PQ\right) + VU \phantom{000} (PY = YQ) \\ & = {1 \over 4}PQ + VU \\ & = {1 \over 4}PQ + {1 \over 2}PQ \phantom{000} \text{(Mid-point theorem)} \\ & = {3 \over 4}PQ \phantom{0} \text{ (Shown)} \end{align}

(ii)

\begin{align} \text{When } & TU = 8, \\ 8 & = {3 \over 4}PQ \\ \\ PQ & = {8 \over {3 \over 4}} \\ & = {32 \over 3} \text{ units} \end{align}

(i)

\begin{align} \angle ADB & = 90^\circ \phantom{000} (\text{Right } \angle \text{ in semicircle)} \\ \\ \angle ADC & = 180^\circ - \angle ADB \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line)} \\ & = 180^\circ - 90^\circ \\ & = 90^\circ \\ & = \angle ADB \\ \\ \angle ADC & = \angle ADB \phantom{000} \text{ [R]} \\ \\ AC & = AB \phantom{00} \text{ (Given) [H]} \\ \\ AD & \text{ is the common side [S]} \\ \\ \\ \therefore \triangle ADC & \text{ is congruent to } \triangle ADB \text{ (RHS)} \end{align}
\begin{align} DC & = DB \phantom{000} (\triangle ADC \equiv \triangle ADB) \\ \\ \therefore D & \text{ is the mid-point of } CB \\ \\ \\ \text{Since } D \text{ is the} & \text{ mid-point of }CB \text{ and } G \text{ is the mid-point of } EB, \\ CE \phantom{.} & // \phantom{.} DG \phantom{000} (\text{Mid-point theorem}) \end{align}

(ii)

\begin{align} \angle FAE & = \angle DAG \phantom{000} \text{ (Common }\angle ) \text{ [A]} \\ \\ \angle AEF & = \angle AGD \phantom{000} \text{ (Corresponding } \angle s) \text{ [A]} \\ \\ \therefore \triangle AEF & \text{ is similar to } \triangle AGD \text{ (AA Similarity)} \end{align}
\begin{align} \text{Since } \triangle AEF & \text{ is similar to } \triangle AGD, \\ {AF \over AD} & = {AE \over AG} \\ {AF \over AD} & = {AE \over 2AE} \phantom{000} (\text{Since } AE = EG, \phantom{.} AG = 2AE) \\ {AF \over AD} & = {1 \over 2} \\ AF & = {1 \over 2}AD \phantom{00} \text{ (Shown)} \end{align}

(i)

\begin{align} \angle DPF & = \angle OPC \phantom{000} \text{ (Common } \angle ) \text{ [A]} \\ \\ \text{Let } & \angle COA = x^\circ \\ \\ \angle POC & = 180^\circ - \angle COA \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line)} \\ & = 180 - x \\ & = (180 - x)^\circ \\ \\ \angle COE & = \angle COA + \angle AOE \\ & = \angle COA + \angle COA \phantom{000} (\angle COA = \angle AOE, \text{ since arc } AC = \text{ arc } AE) \\ & = x + x \\ & = (2x)^\circ \\ \\ \angle CDF & = {1 \over 2}\angle COE \phantom{000} (\angle \text{ at centre } = 2 \angle \text{ at circumference)} \\ & = {1 \over 2} (2x) \\ & = x^\circ \\ \\ \angle PDF & = 180^\circ - \angle CDF \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line)} \\ & = 180 - x \\ & = (180 - x)^\circ \\ & = \angle POC \\ \\ \angle PDF & = \angle POC \phantom{000} \text{ [A]} \\ \\ \\ \therefore \triangle PDF & \text{ is similar to } \triangle POC \text{ (AA Similarity)} \end{align}

(ii)

\begin{align} \text{Since } \triangle PDF & \text{ is similar to } \triangle POC, \\ {PD \over PO} & = {PF \over PC} \\ (PD)(PC) & = (PF)(PO) \\ PD \times PC & = PF \times PO \phantom{00} \text{ (Shown)} \end{align}

(iii)

\begin{align} \text{Let } & \angle OAC = y^\circ \\ \\ \angle CDB & = 180^\circ - \angle OAC \phantom{000} (\angle s \text{ in opposite segment)} \\ & = 180 - y \\ & = (180 - y)^\circ \\ \\ \angle PDB & = 180^\circ - \angle CDB \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line)} \\ & = 180 - (180 - y) \\ & = 180 - 180 + y \\ & = y^\circ \\ & = \angle OAC \\ & = \angle PAC \phantom{000} (\text{Common } \angle) \\ \\ \angle PDB & = \angle PAC \phantom{000} \text{ [A]} \\ \\ \angle DPB & = \angle APC \phantom{000} \text{ (Common } \angle ) \text{ [A]} \\ \\ \\ \therefore \triangle PDB & \text{ is similar to } \triangle PAC \text{ (AA Similarity)} \end{align}
\begin{align} \text{Since } \triangle PDB & \text{ is similar to } \triangle PAC, \\ {PD \over PA} & = {PB \over PC} \\ (PD)(PC) & = (PB)(PA) \\ PD \times PC & = PB \times PA \\ \\ \text{From part (i), } & PD \times PC = PF \times PO. \\ \\ \therefore PF \times PO & = PB \times PA \text{ (Shown)} \end{align}

(i)

\begin{align} \angle QDC & = 90^\circ \phantom{000} \text{(Tangent } \perp \text{ radius)} \\ \\ \angle ADF & = \angle ADQ + \angle QDC + \angle CDF \\ & = \angle ADQ + 90^\circ + \angle CFD \phantom{000} \text{ (Isosceles } \triangle CDF, \angle CDF = \angle CFD) \\ & = \angle ADQ + 90^\circ + \angle OFA \phantom{000} (\text{Common } \angle) \\ & = \angle ADQ + 90^\circ + \angle OAF \phantom{000} \text{ (Isosceles } \triangle OAF, \angle OFA = \angle OAF) \\ & = \angle ADQ + 90^\circ + \angle QAD \phantom{000} (\text{Common } \angle) \\ & = \angle ADQ + 90^\circ + (180^\circ - 90^\circ - \angle ADQ) \phantom{000} (\angle \text{ sum of } \triangle) \\ & = \angle ADQ + 90^\circ + 90^\circ - \angle ADQ \\ & = 180^\circ \\ \\ \therefore A, D & \text{ and } F \text{ lie on the same straight line} \end{align}

(ii)

\begin{align} \angle AFB & = 90^\circ \phantom{000} (\text{Right } \angle \text{ in semicircle}) \\ & = \angle AQD \phantom{00} \text{ [A]} \\ \\ \angle BAF & = \angle DAQ \phantom{000} \text{ (Common } \angle ) \text{ [A]} \\ \\ \therefore \triangle BAF & \text{ is similar to } \triangle DAQ \text{ (AA Similarity)} \end{align}
\begin{align} \text{Since } \triangle BAF & \text{ is similar to } \triangle DAQ, \\ {BA \over DA} & = {AF \over AQ} \\ (BA)(AQ) & = (DA)(AF) \\ (AB)(AQ) & = (AD)(AF) \\ AB \times AQ & = AD \times AF \\ \\ \therefore AD \times AF & = AQ \times AB \phantom{00} \text{ (Shown)} \end{align}

(i)

\begin{align} \angle ABF & = \angle ABD - \angle FBD \\ & = \angle EDB - \angle FBD \phantom{00} (\angle ABD = \angle EDB) \\ & = \angle EDB - \angle FDB \phantom{00} (\angle FBD = \angle FDB) \\ & = \angle EDF \\ \\ \therefore \angle ABF & = \angle EDF \phantom{0} \text{ (Shown)} \end{align}

(ii)

\begin{align} \angle BAF & = \angle DEF \phantom{000} \text{ (Isosceles } \triangle CAE) \text{ [A]} \\ \\ \angle ABF & = \angle EDF \phantom{000} \text{ [A]} \\ \\ BF & = DF \phantom{000} \text{ (Isosceles } \triangle FBD) \text{ [S]} \\ \\ \\ \therefore \triangle ABF & \text{ is congruent to } \triangle EDF \text{ (AAS)} \end{align}

(iii)

\begin{align} AF & = EF \phantom{00} (\triangle ABF \text{ is congruent to } \triangle EDF) \\ \\ \text{Since } & AE = AF + FE, F \text{ is the mid-point of } AE \end{align}

(i)

\begin{align} \angle DYC & = \angle YXB \phantom{000} \text{(Corresponding }\angle s) \text{ [A]} \\ \\ \angle YDC & = \angle AYD \phantom{000} \text{(Alternate }\angle s) \\ & = \angle XYB \phantom{000} \text{(Vertically opposite } \angle s) \text{ [A]} \\ \\ \\ \therefore \triangle YDC & \text{ is similar to } \triangle XYB \text{ (AA Similarity)} \end{align}

(ii)

\begin{align} \text{Since } \triangle YDC & \text{ is similar to } \triangle XYB, \\ {YC \over XB} & = {DC \over YB} \\ {YC \over XB} & = {DC \over {1 \over 2}DC} \phantom{00} (AY = YB) \\ {YC \over XB} & = {1 \over {1 \over 2}} \\ {YC \over XB} & = 2 \\ YC & = 2XB \phantom{0} \text{(Shown)} \end{align}

(i)

\begin{align} \angle AGE & = 180^\circ - \angle CDG \phantom{00} \text{(Interior }\angle s, \phantom{.} CD \phantom{.} // \phantom{.} EG) \\ & = 180^\circ - 90^\circ \\ & = 90^\circ \\ & = \angle ACE \\ \\ \angle AGE & = \angle ACE \phantom{000} \text{ [A]} \\ \\ \angle GAE & = \angle CAE \phantom{000} (AE \text{ bisects } \angle BAC) \text{ [A]} \\ \\ AE & \text{ is the common side [S]} \\ \\ \\ \triangle AGE & \text{ is congruent to } \triangle ACE \text{ (AAS)} \end{align} $$ \therefore GE = CE \phantom{000} (\triangle AGE \equiv \triangle ACE) $$

(ii)

\begin{align} \angle CFE & = \angle FEG \phantom{000} \text{ (Alternate } \angle s) \\ & = \angle AEG \phantom{000} \text{ (Common } \angle ) \\ & = \angle AEC \phantom{000} (\triangle AGE \text{ congruent to } \triangle ACE) \\ & = \angle CEF \phantom{000} \text{ (Common } \angle ) \\ \\ \therefore \triangle CFE & \text{ is an isosceles triangle}. \\ \\ \therefore CF & = CE \phantom{00} \text{ (Isosceles } \triangle CEF) \end{align}

(i)

\begin{align} \text{Since } S \text{ is the } & \text{mid-point of } PQ \text{ and } M \text{ is the mid-point of } PT, \\ \\ SM \phantom{.} & // \phantom{.} QT \phantom{000} (\text{Mid-point theorem}) \end{align}
\begin{align} \angle URT & = \angle SRM \phantom{00} \text{(Common } \angle ) \text{ [A]} \\ \\ \angle UTR & = \angle SMR \phantom{00} \text{(Corresponding } \angle s, SM \phantom{0} // \phantom{0} UT) \text{ [A]} \\ \\ \\ \triangle URT & \text{ is similar to } \triangle SRM \text{ (AA Similarity)} \end{align}
\begin{align} PT & = 2TR \phantom{000} \text{ (Given)} \\ PM + MT & = 2TR \\ PM + PM & = 2TR \phantom{000} (PM = MT) \\ 2PM & = 2TR \\ PM & = TR = MT \end{align}
\begin{align} \text{Since } \triangle URT & \text{ is similar to } \triangle SRM, \\ {SM \over UT} & = {MR \over TR} \\ {SM \over UT} & = {2TR \over TR} \phantom{000} (MT = TR) \\ {SM \over UT} & = 2 \\ SM & = 2UT \\ \\ \therefore SM & = 2TU \phantom{0} \text{(Shown)} \end{align}

(ii)

\begin{align} SM & = {1 \over 2}QT \phantom{000} \text{(Mid-point theorem)} \\ 2TU & = {1 \over 2}QT \phantom{000} (SM = 2TU \text{ from part i}) \\∂ 4TU & = QT \phantom{0} \text{(Shown)} \end{align}

(iii)

\begin{align} \require{cancel} {\text{Area of }\triangle UTR \over \text{Area of }\triangle TQR} & = {\cancel{1 \over 2} \times TU \times \cancel{\text{Height}} \over \cancel{{1 \over 2}} \times QT \times \cancel{\text{Height}}} \\ & = {TU \over QT} \\ & = {\cancel{TU} \over 4\cancel{TU}} \phantom{000} (4TU = QT \text{ from part ii}) \\ & = {1 \over 4} \\ \\ {\text{Area of }\triangle UTR \over \text{Area of }\triangle TQR} & = {1 \over 4} \\ \\ 4 \times \text{Area of }\triangle UTR & = \text{Area of }TQR \\ \\ \\ \text{Since area of } & \triangle UTR = x, \\ \\ \text{Area of }TQR & = 4 \times x \\ & = 4x \text{ cm}^2 \end{align}

(i)

(Consider triangle ABY) \begin{align} \text{Since } E \text{is the mid-point} & \text{ of } AB \text{ and } X \text{ is the mid-point of } AY, \\ \\ EX \phantom{0} & // \phantom{0} BY \phantom{000} \text{ (Mid-point theorem)} \\ \\ ED \phantom{0} & // \phantom{0} BY \phantom{000} (EXD \text{ is a straight line}) \\ \\ \therefore XD \phantom{0} & // \phantom{0} BY \end{align}
(Consider triangle CXB) \begin{align} \text{Since } Y \text{ is the mid-point} & \text{ of } XC \text{ and } F \text{ is the mid-point of } BC, \\ \\ FY \phantom{0} & // \phantom{0} BX \phantom{000} \text{ (Mid-point theorem)} \\ \\ FD \phantom{0} & // \phantom{0} BX \phantom{000} (FYD \text{ is a straight line}) \\ \\ \therefore YD \phantom{0} & // \phantom{0} BX \\ \\ \\ \text{Since } XD \phantom{.} // \phantom{.} BY & \text{ and } YD \phantom{.} // \phantom{.} BX, \phantom{.} BXDY \text{ is a parallelogram} \end{align}

(ii)

\begin{align} AX & = CY \phantom{000} \text{ (Given) [S]} \\ \\ \angle AXB & = 180^\circ - \angle BXY \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line)} \\ & = 180^\circ - \angle XYD \phantom{000} (\text{Alternate } \angle s \text{, } BX \phantom{0} // \phantom{0} YD) \\ & = \angle CYD \\ \\ \angle AXB & = \angle CYD \phantom{000} \text{ [A]} \\ \\ XB & = YD \phantom{000} \text{ (Opposite sides of parallelogram } BXDY) \text{ [S]} \\ \\ \\ \therefore \triangle AXB & \text{ is congruent to } \triangle CYD \text{ (SAS)} \end{align}

(iii)

\begin{align} AX & = CY \phantom{000} \text{ (Given) [S]} \\ \\ \angle AXD & = 180^\circ - \angle DXY \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line)} \\ & = 180^\circ - \angle BYX \phantom{000} (\text{Alternate } \angle s \text{, } XD \phantom{0} // \phantom{0} BY) \\ & = \angle CYB \\ \\ \angle AXD & = \angle CYB \phantom{000} \text{ [A]} \\ \\ XD & = YB \phantom{000} \text{ (Opposite sides of parallelogram } BXDY) \text{ [S]} \\ \\ \\ \therefore \triangle AXD & \text{ is congruent to } \triangle CYB \text{ (SAS)} \\ \\ \\ AB & = DC \phantom{000} (\triangle AXB \equiv \triangle CYD) \\ \\ AD & = BC \phantom{000} (\triangle AXD \equiv \triangle CYB) \\ \\ \\ \text{Since } & AB = DC \text{ and } AD = BC, \phantom{.} ABCD \text{ is a parallelogram} \end{align}

\begin{align} \text{Let } & \angle BDA = x^\circ \\ \\ \angle DBA & = \angle BDA \phantom{000} \text{(Isosceles } \triangle ADB) \\ \\ \angle BAD & = 180^\circ - \angle DBA - \angle BDA \phantom{000} (\angle \text{ sum of } \triangle) \\ & = 180 - x - x \\ & = (180 - 2x)^\circ \\ \\ \angle BCD & = 180^\circ - \angle BAD \phantom{000} (\angle s \text{ in opposite segment)} \\ & = 180 - (180 - 2x) \\ & = 180 - 180 + 2x \\ & = 2x \\ & = 2(x) \\ & = 2\angle BDA \\ \\ \therefore \angle BCD & = 2\angle BDA \phantom{0} \text{ (Shown)} \end{align}

\begin{align} \text{Let } & \angle PBC = x^\circ \\ \\ \angle CQP & = \angle PBC \phantom{000} (\angle s \text{ in the same segment)} \\ & = x^\circ \\ \\ \angle QRP & = \angle CQP \phantom{000} \text{(Alternate segment theorem)} \\ & = x^\circ \\ \\ \angle QRB & = \angle QRP \phantom{000} \text{(Common } \angle ) \\ & = x^\circ \\ & = \angle PBC \\ & = \angle RBC \phantom{000} \text{(Common } \angle ) \\ \\ \therefore \angle QRB & = \angle RBC \\ \\ \therefore BC \phantom{.} & // \phantom{.} QR \phantom{000} \text{(Converse of alternate } \angle s) \end{align}

(i)

\begin{align} PQ^2 & = PA \times PB \phantom{000000} [\text{Tangent secant theorem}] \\ \\ \text{Since tangents meet at } & P, \phantom{.} PQ = PR \\ \\ \text{Since } H \text{ is midpoint of } & QR, \phantom{.} \angle PHQ = 90^\circ \\ \\ PC^2 & = PH^2 + HC^2 \\ & = PH^2 + (QH - QC)^2 \\ & = \underbrace{PH^2}_\text{Need to find this} + QH^2 - 2(QH)(QC) + QC^2 \\ \\ \\ PQ^2 & = PH^2 + QH^2 \\ PA \times PB & = PH^2 + QH^2 \\ \\ PH^2 & = PA \times PB - QH^2 \\ \\ \\ \therefore PC^2 & = PA \times PB - QH^2 + QH^2 - 2(QH)(QC) + QC^2 \\ & = PA \times PB - 2(QH)(QC) + QC^2 \\ & = PA \times PB + QC(QC - 2QH) \\ & = PA \times PB - QC(2QH - QC) \\ & = PA \times PB - QC(QR - QC) \\ & = PA \times PB - QC(RC) \\ & = PA \times PB - QC \times RC \\ & = PA \times PB - AC \times BC \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} PC^2 & = PA \times PB - AC \times BC \\ PC^2 & = PA \times PB - (PC - PA)(PB - PC) \\ PC^2 & = PA \times PB - [(PB)(PC) - PC^2 - (PA)(PB) + (PA)(PC)] \\ PC^2 & = PA \times PB - (PB)(PC) + PC^2 + (PA)(PB) - (PA)(PC) \\ 0 & = 2(PA)(PB) - (PB)(PC) - (PA)(PC) \\ (PB)(PC) + (PA)(PC) & = 2(PA)(PB) \\ PC(PB + PA) & = 2(PA)(PB) \\ {PB + PA \over 2(PA)(PB)} & = {1 \over PC} \\ {1 \over 2} \left[{PB + PA \over (PA)(PB)}\right] & = {1 \over PC} \\ {1 \over 2} \left[ {PB \over (PA)(PB)} + {PA \over (PA)(PB)} \right] & = {1 \over PC} \\ {1 \over 2} \left({1 \over PA} + {1 \over PB} \right) & = {1 \over PC} \phantom{0000} \text{(Shown)} \end{align}