(i)

\begin{align} AD & = CD \phantom{0} \text{ (}D \text{ is mid-point of }AC\text{)} \text{ [S]} \\ \\ \angle ADB & = \angle CDB = 90^\circ \phantom{0} \text{(} \angle\text{s on a straight line)} \text{ [A]} \\ \\ BD & \text{ is a common length} \text{ [S]} \\ \\ \\ \therefore \triangle ABD & \text{ is congruent to } \triangle CBD \text{ (SAS)} \end{align}

(ii)

\begin{align} \text{Since } \triangle ABD & \text{ is congruent to } \triangle CBD, \\ \angle ABD & = \angle CBD \\ \\ \therefore BD & \text{ bisects } \angle ABC \end{align}

(i)

\begin{align} AP & = EQ \phantom{0} \text{ (Given)} \\ \\ \therefore AQ & = AP + PQ \\ & = EQ + PQ \\ & = EP \text{ (Shown)} \end{align}

(ii)

\begin{align} \angle CPQ & = \angle CQP \phantom{0} \text{(Isosceles triangle since } CP = CQ) \\ \\ \therefore \angle AQB & = \angle EPD \phantom{0} \text{ [A]} \\ \\ AQ & = EP \phantom{0} \text{ (From part i)} \text{ [S]} \\ \\ \angle BAQ & = \angle DEP \phantom{0} \text{ (Given)} \text{ [A]} \\ \\ \\ \therefore \triangle ABQ \text{ is } & \text{congruent to } \triangle EDP \text{ (ASA)} \end{align}

(iii)

\begin{align} \text{Since } \triangle ABQ & \text{ is congruent to } \triangle EDP, \\ \\ \angle QBA & = \angle PDE \end{align}

(i)

\begin{align} \text{Since } \triangle BOA & \text{ is similar to } \triangle BYT, \\ {AB \over TB} & = {BO \over BY} \\ (AB)(BY) & = (TB)(BO) \\ {AB \over BO} & = {TB \over BY} \\ {AB \over BO} & = {BT \over BY} \text{ (Shown)} \end{align}

(ii)

\begin{align} {AB \over BO} & = {BT \over BY} \phantom{0} \text{ (From part i)} \text{ [S,S]} \\ \\ \angle ABT & = \angle OBY \phantom{0} \text{ (Vertically opposite } \angle {s)} \text{ [A]} \\ \\ \\ \therefore \triangle BAT & \text{ is similar to } \triangle BOY \text{ (SAS Similarity)} \end{align}

(i)

\begin{align} \angle BAC & = \angle MAN \phantom{0} \text{ (Common } \angle \text{)} \text{ [A]} \\ \\ \angle ABC & = \angle AMN \phantom{0} \text{ (Corresponding } \angle \text{s)} \text{ [A]} \\ \\ \\ \therefore \triangle ABC & \text{ is similar to } \triangle AMN \text{ (AA Similarity)} \end{align}

(ii)

\begin{align} \text{Since } \triangle ABC & \text{ is similar to } \triangle AMN, \\ {MN \over BC} & = {AN \over AC} \end{align} \begin{align} {MN \over BC} & = {AN \over AC} \\ {NC \over BC} & = {AN \over AC} \phantom{00000} (MN = NC)\\ (NC)(AC) & = (AN)(BC) \\ NC \times AC & = AN \times BC \text{ (Shown)} \end{align}

(i)

\begin{align} \text{Let } \angle & CBD = x^\circ \\ \\ \angle CBD & = 180 - 90 - x \phantom{0} \text{ (} \angle \text{ sum of triangle)} \\ & = (90 - x)^\circ \\ \\ \angle ABC & = 90 - (90 - x) \\ & = 90 - 90 + x \\ & = x^\circ \\ \\ \therefore \angle CDB & = \angle ABC = x^\circ \phantom{0} \text{ [A]} \\ \\ \angle BCD & = \angle CAB = 90^\circ \phantom{0} \text{ [A]} \\ \\ \\ \therefore \triangle ABC & \text{ is similar to } \triangle CDB \text{ (AA Similarity)} \end{align}

(ii)

\begin{align} \text{Since } \triangle ABC & \text{ is similar to } \triangle CDB, \\ {AC \over CB} & = {CB \over BD} \\ (AC)(BD) & = (CB)(CB) \\ AC \times BD & = CB^2 \text{ (Shown)} \end{align}

\begin{align} \text{Since } AG = GB \text{ and } & CF = FB, \text{ by Mid-point theorem,} \\ \\ GF \phantom{0} &// \phantom{0} AC \\ \\ \therefore AC \phantom{0} & // \phantom{0} GE \\ \\ \\ \text{Since } GE \phantom{0} & // \phantom{0} BD, \\ AC \phantom{0} & // \phantom{0} BD \text{ (Shown)} \end{align}

(i)

\begin{align} \text{Since } AD = DB \text{ and } AE = EC, \phantom{.} D & \text{ is the mid-point of }AB \text{ and } E \text{ is the mid-point of } AC.\\ \\ \text{By Mid-} & \text{point theorem, } DE \phantom{0} // \phantom{0} BC \end{align}

(ii)

\begin{align} \text{By Mid} & \text{-point theorem,} \\ DE & = {1 \over 2} BC \\ & = {1 \over 2}(2x) \\ & = x \text{ cm} \end{align}

(i)

\begin{align} \angle SDM & = \angle TCM \phantom{000} \text{ (Given) [A]} \\ \\ DM & = CM \phantom{000} \text{ (} M \text{ is the midpoint of }DC) \text{ [S]} \\ \\ \angle DMS & = \angle CMT \phantom{000} \text{ (Vertically opposite } \angle s) \text{ [A]} \\ \\ \\ \therefore \triangle SDM & \text{ is congruent to } \triangle TCM \text{ (ASA)} \end{align}

(ii)

\begin{align} \angle DSM & = \angle CTM \phantom{000} (\triangle SDM \equiv \triangle TCM) \\ \\ \therefore \angle DSB & = \angle CTA \phantom{000} \text{ [A]} \\ \\ \\ SD & = TC \phantom{000} (\triangle SDM \equiv \triangle TCM) \text{ [S]} \\ \\ \\ \angle SDM & = \angle TCM \phantom{000} (\triangle SDM \equiv \triangle TCM) \\ \\ 2\angle SDB & = 2\angle TCA \phantom{000} (DB \text{ bisects } \angle SDM \text{ and } CA \text{ bisects } \angle TCM) \\ \\ \angle SDB & = \angle TCA \phantom{000} \text{ [A]} \\ \\ \\ \therefore \triangle SDB & \equiv \triangle TCA \text{ (ASA)} \\ \\ \therefore SB & = TA \end{align}

\begin{align} \angle GAE & = \angle GEA \phantom{000} (\text{Isosceles } \triangle \text{ since } AG = EG) \\ \\ \therefore \angle NAT & = \angle LER \phantom{000} \text{ [A]} \end{align} \begin{align} AN & = AG - NG \\ & = EG - NG \phantom{000} (AG = EG) \\ & = EG - LG \phantom{000} (NG = LG) \\ & = EL \\ \\ \therefore AN & = EL \phantom{000} \text{ [S]} \end{align} \begin{align} TI & = {1 \over 2}NI \phantom{000} (T \text{ is midpoint of } NI) \\ & = {1 \over 2} LI \phantom{000} (NI = LI)\\ & = RI \phantom{000} (R \text{ is the mid-point of } LI) \end{align} \begin{align} \angle ATN & = \angle ITR \phantom{000} \text{ (Vertically opposite }\angle s) \\ & = \angle IRT \phantom{000} (\text{Isosceles } \triangle ITR) \\ & = \angle ERL \phantom{000} \text{ (Vertically opposite }\angle s) \\ \\ \angle ATN & = \angle ERL \phantom{000} \text{ [A]} \end{align} $$ \therefore \triangle ANT \equiv \triangle ELR \text{ (AAS)} $$

\begin{align} JQ & = FX \phantom{000} (\text{Given}) \text{ [S]} \\ \\ \angle JQZ & = \angle FXZ \phantom{000} (\text{Given}) \text{ [A]} \\ \\ QZ & = XZ \phantom{000} (\text{Given}) \text{ [S]} \\ \\ \\ \therefore \triangle QJZ & \equiv \triangle XFZ \text{ (SAS)} \end{align} \begin{align} \angle QJZ & = \angle XFZ \phantom{000} (\triangle JQZ \equiv \triangle FXZ) \\ \\ \therefore \angle ZJX & = \angle ZFQ \phantom{000} \text{ [A]} \\ \\ \\ JX & = JQ + QX \\ & = XF + QX \phantom{000} (JQ = XF) \\ & = FQ \\ \\ \therefore JX & = FQ \phantom{000} \text{ [S]} \\ \\ \\ \angle ZXQ & = \angle ZQX \phantom{000} (\text{Isosceles } \triangle) \\ \\ \therefore \angle ZXJ & = \angle ZQF \phantom{000} \text{ [A]} \\ \\ \\ \therefore \triangle JZX & \equiv \triangle FZQ \text{ (ASA)} \end{align}

(i)

\begin{align} \angle BAE & = \angle DAC \phantom{000} (AD \text{ bisects } \angle BAD) \text{ [A]} \\ \\ \\ \angle DCE & = \angle DEC \phantom{000} (\text{Isosceles } \triangle CDE) \\ & = \angle AEB \phantom{000} (\text{Vertically opposite } \angle s) \\ \\ \therefore \angle AEB & = \angle ACD \phantom{000} \text{ [A]} \\ \\ \\ \therefore \triangle ABE & \text{ is similar to } \triangle ADC \text{ (AA Similarity)} \end{align}

(ii)

\begin{align} \text{Since } \triangle ABE & \text{ is similar to } \triangle ADC, \\ {AB \over AD} & = {AE \over AC} \\ (AB)(AC) & =(AD)(AE) \\ AB \times AC & = AD \times AE \text{ (Shown)} \end{align}

(i)

\begin{align} \angle EFD & = \angle FEC \phantom{000} \text{ (Alternate } \angle s) \\ & = 90^\circ \\ & = \angle BDE \\ \\ \angle EDF & = \angle BED \phantom{000} \text{ [A]} \\ \\ \\ \angle EDF & = \angle BED \phantom{000} \text{ (Alternate } \angle s) \text{ [A]} \\ \\ \\ \therefore \triangle BDE & \text{ is similar to } \triangle EFD \text{ (AA Similarity)} \end{align}

(ii)

\begin{align} \angle EFD & = \angle CEF = 90^\circ \phantom{000} \text{ [A]} \\ \\ \\ \text{Let } & \angle ABC = x^\circ \\ \\ \angle DBE & = \angle ABC \\ & = x^\circ \\ \\ \angle DEB & = 180^\circ - 90^\circ - x^\circ \phantom{000} (\angle \text{ sum of triangle)} \\ & = (90 - x)^\circ \\ \\ \angle FED & = 180^\circ - 90^\circ - (90 - x)^\circ \phantom{000} (\angle s \text{ on a straight line)} \\ & = 90 - 90 + x \\ & = x^\circ \\ \\ \angle ECF & = \angle ACB \\ & = \angle ABC \phantom{000} \text{ (Isosceles } \triangle ABC) \\ & = x^\circ \\ & = \angle FED \\ \\ \therefore \angle ECF & = \angle FED \phantom{000} \text{ [A]} \\ \\ \\ \therefore \triangle EFD & \text{ is similar to } \triangle CEF \text{ (AA Similarity)} \end{align}

(i)

\begin{align} \text{Let } D \text{ denote} & \text{ the point where } BE \text{ and } FA \text{ meets.} \\ \\ \angle DBF & = \angle DAE \phantom{000} \text{ (Given) [A]} \\ \\ \angle BDF & = \angle ADE \phantom{000} \text{ (Vertically opposite }\angle s) \text{ [A]} \\ \\ \therefore \triangle DBF & \text{ is similar to } \triangle DAE \text{ (AA Similarity)} \end{align} \begin{align} \angle DFB & = \angle DEA \phantom{000} (\triangle DBF \text{ is similar to } \triangle DAE) \\ \\ \therefore \angle AFC & = 180^\circ - \angle DFB \phantom{000} \text{(} \angle s \text{ on a straight line)} \\ & = 180^\circ - \angle DEA \phantom{000} (\angle DFB = \angle DEA) \\ & = \angle BEC \text{ (Shown)} \end{align}

(ii)

\begin{align} \angle AFC & = \angle BEC \phantom{000} \text{ (From part i) [A]} \\ \\ \angle ACF & = \angle BCE \phantom{000} \text{ (Common angle) [A]} \\ \\ \therefore \triangle AFC & \text{ is similar to } \triangle BEC \text{ (AA Similarity)} \end{align}

(iii)

\begin{align} \text{Since } \triangle AFC & \text{ is similar to } \triangle BEC, \\ {CF \over CE} & = {CA \over CB} \\ (CF)(CB) & = (CE)(CA) \\ CF \times CB & = CE \times CA \text{ (Shown)} \end{align}

(i)

\begin{align} \angle BAC & = \angle EDC = 90^\circ \phantom{000} \text{ (Given) [A]} \\ \\ \angle BCA & = \angle ECD \phantom{000} \text{ (Common } \angle) \text{ [A]} \\ \\ \\ \therefore \triangle BAC & \text{ is similar to } \triangle EDC \text{ (AA Similarity)} \end{align} \begin{align} \require{cancel} \text{Since } \triangle BAC & \text{ is similar to } \triangle EDC, \\ {DC \over AC} & = {EC \over BC} \\ {DC \over AC} & = {\cancel{EC} \over 2\cancel{EC}} \phantom{000} (E \text{ is the mid-point of } BC) \\ {DC \over AC} & = {1 \over 2} \\ 2DC & = AC \end{align} \begin{align} AC & = AD + DC \\ 2DC & = AD + DC \phantom{000} (AC = 2DC) \\ 2DC - DC & = AD \\ DC & = AD \\ \\ \therefore D \text{ is the} & \text{ mid-point of } AC. \end{align}

(ii) Consider triangles EDA and EDC

\begin{align} DA & = DC \phantom{000} \text{ (From part i) [S]} \\ \\ \\ \angle EDA & = 180^\circ - 90^\circ \phantom{000} (\angle s \text{ on a straight line}) \\ & = 90^\circ \\ & = \angle EDC \\ \\ \angle EDA & = \angle EDC \phantom{000} \text{ [A]} \\ \\ \\ ED & \text{ is a common side [S]} \\ \\ \\ \therefore \triangle EDA & \text{ is congruent to } \triangle EDC \text{ (SAS)} \end{align} \begin{align} \angle EAD & = \angle ECD \phantom{000} (\triangle EDA \equiv \triangle EDC) \\ \\ \therefore \triangle AEC & \text{ is an isosceles triangle} \end{align}

(i)

\begin{align} \angle DBG & = \angle ABC \phantom{000} (\text{Common } \angle) \text{ [A]} \\ \\ \angle DGB & = \angle ACB \phantom{000} (\text{Corresponding } \angle s) \text{ [A]} \\ \\ \\ \triangle DBG & \text{ is similar to } \triangle ABC \text{ (AA Similarity}) \end{align} \begin{align} \text{Since } \triangle DBG & \text{ is similar to } \triangle ABC, \\ \\ {DB \over AB} & = {DG \over AC} \\ \\ {CE \over AB} & = {DG \over AC} \phantom{000} (DB = CE) \\ \\ {CE \over AC} & = {DG \over AC} \phantom{000} (\text{Isosceles } \triangle ABC, AB = AC) \\ \\ \therefore CE & = DG \text{ (Shown)} \end{align}

(ii)

\begin{align} \angle DFG & = \angle EFC \phantom{000} \text{ (Vertically opposite }\angle s) \text{ [A]} \\ \\ \angle GDF & = \angle CEF \phantom{000} \text{ (Alternate }\angle s) \text{ [A]} \\ \\ DG & = EC \phantom{000} \text{ (From part i) [S]} \\ \\ \\ \therefore \triangle DGF & \text{ is congruent to } \triangle ECF \text{ (AAS)} \end{align}

(iii)

\begin{align} DF & = FE \phantom{000} (\triangle DGF \equiv \triangle ECF) \end{align}

(i)

$$ AB = QC \phantom{0} \text{ (Given) [S]} $$ \begin{align} \text{Let } & \angle ABP \text{ be } x^\circ. \\ \\ \angle BPF & = 180^\circ - 90^\circ - x^\circ \phantom{000} (\angle \text{ sum of triangle)} \\ & = (90 - x)^\circ \\ \\ \angle EPC & = \angle BPF \phantom{000} \text{ (Vertically opposite } \angle s) \\ & = (90 - x)^\circ \\ \\ \angle PEC & = 90^\circ \\ \\ \angle QCA & = \angle PCE \phantom{000} (\angle \text{ sum of triangle)} \\ & = 180^\circ - 90^\circ - (90 - x)^\circ \\ & = 90 - 90 + x \\ & = x^\circ \\ \\ \therefore \angle ABP & = \angle QCA \phantom{000} \text{ [A]} \end{align} \begin{align} BP & = CA \phantom{0} \text{ (Given) [S]} \\ \\ \\ \therefore \triangle ABP & \text{ is congruent to } \triangle QCA \text{ (SAS)} \end{align}

(ii)

\begin{align} AP & = QA \phantom{000} (\triangle ABP \equiv \triangle QCA) \\ \\ \therefore \triangle AQP & \text{ is an isosceles } \triangle \\ \\ \\ \angle BAP & = \angle CQA \phantom{000} (\triangle ABP \equiv \triangle QCA) \\ & = \angle PQA \phantom{000} (\text{Common } \angle) \\ & = \angle APQ \phantom{000} (\text{Isosceles } \triangle AQP) \\ \\ \therefore \angle BAP & = \angle APQ \text{ (Shown)} \end{align}

(iii)

\begin{align} \text{Let } \angle & BAP \text{ be } y^\circ. \\ \\ \angle APQ & = \angle BAP \phantom{000} (\text{From part ii}) \\ & = y^\circ \\ \\ \angle PQA & = \angle APQ \phantom{000} (\text{Isosceles } \triangle AQP) \\ & = y^\circ \\ \\ \angle QFA & = \angle BFP \phantom{000} \text{ (Vertically opposite } \angle s) \\ & = 90^\circ \\ \\ \angle QAF & = 180^\circ - 90^\circ - y^\circ \phantom{000} (\angle \text{ sum of triangle)} \\ & = (90 - y)^\circ \\ \\ \angle QAP & = (90 - y)^\circ + y^\circ \\ & = 90^\circ \\ \\ \therefore AP & \text{ is perpendicular to } AQ. \end{align}

(i)

\begin{align} \angle BAC & = \angle FEC \phantom{000} (\text{Corresponding } \angle s) \text{ [A]} \\ \\ \angle ACB & = \angle ECF \phantom{000} (\text{Common } \angle) \text{ [A]} \\ \\ \\ \triangle BAC & \text{ is similar to } \triangle FEC \text{ (AA Similarity)} \end{align} \begin{align} \text{Since } \triangle BAC & \text{ is similar to } \triangle FEC, \\ {EC \over AC} & = {FC \over BC} \\ {1 \over 2} & = {FC \over BC} \phantom{000} (E \text{ is the mid-point of } AC) \\ BC & = 2FC \\ \\ \therefore F \text{ is the} & \text{ mid-point of } BC \\ \\ \therefore BF & : BC \\ 1 & : 2 \end{align}

(ii)

\begin{align} \text{Since } E \text{ is the mid} & \text{-point of } AC \text{ and } F \text{ is the mid-point of } BC, \\ EF & = {1 \over 2} AB \phantom{000} (\text{Mid-point theorem}) \\ \\ \therefore AB & : EF \\ 2 & : 1 \end{align} \begin{align} \angle AEF & = \angle ACD \phantom{000} (\text{Corresponding } \angle s) \text{ [A]} \\ \\ \angle FAE & = \angle DAC \phantom{000} (\text{Common } \angle ) \text{ [A]} \\ \\ \\ \triangle AEF & \text{ is similar to } \triangle ACD \text{ (AA Similarity)} \end{align} \begin{align} \text{Since } \triangle AEF & \text{ is similar to } \triangle ACD, \\ \\ {AE \over AC} & = {EF \over CD} \\ \\ {1 \over 2} & = {EF \over CD} \phantom{000} (E \text{ is the mid-point of } AC) \\ \\ CD & = 2EF \\ \\ \therefore CD & : EF \\ 2 & : 1 \end{align} \begin{align} AB : E & F : CD \\ 2 : \phantom{E} & 1 \phantom{.} : 2 \\ \\ \therefore AB & : CD \\ 2 & : 2 \\ 1 & : 1 \text{ (Shown)} \end{align}

(i)

\begin{align} EC & = 2EF \phantom{000} (F \text{ is the mid-point of } EC) \\ \\ AE & = AC - EC \\ & = 3EC - EC \phantom{000} (AC = 3EC) \\ & = 2EC \\ & = 2(2EF) \phantom{00000.} (EC = 2EF) \\ & = 4EF \text{ (Shown)} \end{align}

(ii) Consider triangle CBE, where D is the midpoint of BC and F is the midpoint of EC

\begin{align} BE \phantom{.} & // \phantom{.} DF \phantom{000} (\text{Mid-point theorem}) \\ \\ \\ \angle AGE & = \angle ADF \phantom{000} (BE \phantom{.} // \phantom{.} DF, \text{ Corresponding } \angle s) \text{ [A]} \\ \\ \angle GAE & = \angle DAF \phantom{000} (\text{Common } \angle) \text{ [A]} \\ \\ \triangle AGE & \text{ is similar to } \triangle ADF \text{ (AA Similarity)} \end{align} \begin{align} \text{Since } \triangle AGE & \text{ is similar to } \triangle ADF, \\ {AG \over AD} & = {AE \over AF} \\ {AG \over AD} & = {AE \over AE + EF} \\ {AG \over AD} & = {4EF \over 4EF + EF} \phantom{000} \text{(From part i, } AE = 4EF) \\ {AG \over AD} & = {4EF \over 5EF} \\ {AG \over AD} & = {4 \over 5} \\ {AG \over AG + GD} & = {4 \over 5} \\ 5(AG) & = 4(AG + GD) \\ 5AG & = 4AG + 4GD \\ 5AG - 4AG & = 4GD \\ AG & = 4GD \\ {AG \over GD} & = 4 \text{ (Shown)} \end{align}

\begin{align} \text{Triangles } ABC \text{ and } & EFC \text{ are similar} \\ \\ \text{Since } AB = BC, & \phantom{.} EF = FC \\ \\ \\ \text{Let } \angle BAD & = \theta \\ \\ \angle ABE & = 180^\circ - 90^\circ - \theta \phantom{000000} [\text{Angle sum of triangle}] \\ & = 90^\circ - \theta \\ \\ \angle FBE & = 90^\circ - (90^\circ - \theta) \\ & = \theta \\ & = \angle BAD \phantom{0000000} [A] \\ \\ \angle ABD & = \angle BFE = 90^\circ \phantom{00} [A] \\ \\ \text{Triangles } BAD \text{ and } & FBE \text{ are similar (AA)} \\ \\ {BD \over FE} & = {BA \over FB} \\ {0.5 \over FE} & = {1 \over 1 - FC} \\ 0.5(1 - FC) & = FE \\ 0.5 - 0.5 FC & = FE \\ \\ \text{Since } & EF = FC, \\ 0.5 - 0.5 EF & = EF \\ 0.5 & = 1.5 EF \\ {0.5 \over 1.5} & = EF \\ {1 \over 3} & = EF \\ \\ EF & = {1 \over 3} \text{ cm} \end{align}

(i)

\begin{align} \text{Let } & \angle ABD = \theta \\ \\ \angle BAD & = 180^\circ - 90^\circ - \theta \phantom{000000} [\text{Angle sum of triangle}] \\ & = 90^\circ - \theta \\ \\ \angle ACD & = 90^\circ - (90^\circ - \theta) \\ & = 90^\circ - 90^\circ + \theta \\ & = \theta \\ & = \angle BAD \phantom{0000000} [A] \\ \\ \angle ADB & = \angle CDA = 90^\circ \phantom{00} [A] \\ \\ \\ \text{Triangles } & ABD \text{ and } CAD \text{ are similar (AA)} \end{align}

(ii)

\begin{align} \text{Triangles } & ABD \text{ and } CAD \text{ are similar} \\ \\ {BD \over AD} & = {AB \over CA} \\ BD & = AD \left(AB \over AC\right) \\ \\ {AD \over CD} & = {AB \over CA} \\ AD & = CD \left(AB \over AC\right) \\ \\ \therefore BD & = CD \left(AB \over AC\right) \left(AB \over AC\right) \\ {BD \over CD} & = \left(AB \over AC\right)^2 \\ {BD \over DC} & = {AB^2 \over AC^2} \phantom{00} \text{ (Shown)} \end{align}