(i) Since AC and BD bisect each other at E, E is the midpoint of BD and AC

\begin{align} AE & = CE \phantom{000} (E \text{ is the mid-point of } AC) \text{ [S]} \\ \\ \angle AEB & = \angle CED \phantom{000} \text{ (Vertically opposite }\angle s) \text{ [A]} \\ \\ BE & = DE \phantom{000} (E \text{ is the mid-point of } BD) \text{ [S]} \\ \\ \\ \therefore \triangle ABE & \text{ is congruent to } \triangle CDE \text{ (SAS)} \end{align}

(ii)

\begin{align} \angle ABE & = \angle CDE \phantom{000} (\triangle ABE \equiv \triangle CDE) \\ \\ \angle ABD & = \angle CDB \\ \\ \therefore AB \phantom{.} & // \phantom{.} CD \phantom{000} (\text{Converse of alternate } \angle s) \\ \\ \\ BE & = DE \phantom{000} \text{ [S]} \\ \\ \angle BEC & = \angle DEA \phantom{000} (\text{Vertically opposite } \angle s) \text{ [A]} \\ \\ EC & = EA \phantom{000} \text{ [S]} \\ \\ \therefore \triangle BEC & \text{ is congruent to } \triangle DEA \text{ (SAS)} \\ \\ \\ \angle EBC & = \angle EDA \phantom{000} (\triangle BEC \equiv \triangle DEA) \\ \\ \angle DBC & = \angle BDA \\ \\ \therefore AD \phantom{.} & // \phantom{.} BC \phantom{000} (\text{Converse of alternate } \angle s) \\ \\ \\ \text{Since } AB \phantom{.} // & \phantom{.} DC \text{ and } AD \phantom{.} // \phantom{.} BC, \phantom{.} ABCD \text{ is a parallelogram} \end{align}

(i) Note: AFCE is a square, thus interior angles are right angles

\begin{align} AF & = CE \phantom{000} (AFCE \text{ is a square)} \text{ [S]} \\ \\ \\ \angle AFB & = 180^\circ - 90^\circ \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line}) \\ & = 90^\circ \\ \\ \angle CED & = 180^\circ - 90^\circ \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line}) \\ & = 90^\circ \\ \\ \therefore \angle AFB & = \angle CED \phantom{000} \text{ [A]} \\ \\ \\ BF & = DE \phantom{000} \text{ (Given) [S]} \\ \\ \\ \triangle ABF & \text{ is congruent to } \triangle CDE \text{ (SAS)} \end{align}

(ii)

\begin{align} \text{Let } & \angle ABF \text{ be } x^\circ. \\ \\ \angle BAF & = 180^\circ - 90^\circ - x^\circ \phantom{000} (\angle \text{ sum of triangle)} \\ & = (90 - x)^\circ \\ \\ \angle ABF + \angle BAE & = x^\circ + (90 - x)^\circ + 90^\circ \\ & = 180^\circ \\ \\ \therefore AD \phantom{.} & // \phantom{.} BC \text{ (Converse of interior } \angle s) \\ \\ \\ \angle ABF & = \angle CDE \phantom{000} (\triangle ABF \equiv \triangle CDE) \\ & = x^\circ \\ \\ \angle BAF & = \angle DCE \phantom{000} (\triangle ABF \equiv \triangle CDE) \\ & = (90 - x)^\circ \\ \\ \angle CDE + \angle DCE & = x^\circ + (90 - x)^\circ + 90^\circ \\ & = 180^\circ \\ \\ \therefore AB \phantom{.} & // \phantom{.} DC \text{ (Converse of interior } \angle s) \\ \\ \\ \text{Since } AD \phantom{.} // & \phantom{.} BC \text{ and } AB \phantom{.} // \phantom{.} DC, \phantom{.} ABCD \text{ is a parallelogram} \end{align}

(i)

\begin{align} \angle ADB & = \angle CBD \phantom{000} \text{ (Alternate }\angle s) \text{ [A]} \\ \\ BD & \text{ is the common length} \phantom{000} \text{ [S]} \\ \\ \angle ABD & = \angle CDB \phantom{0} \text{ (Alternate }\angle s) \text{ [A]} \\ \\ \\ \therefore \triangle ABD & \text{ is congruent to } \triangle CDB \text{ (ASA)} \end{align}

(ii)

\begin{align} AD & = CB \phantom{000} (\triangle ABD \equiv \triangle CDB) \\ \\ AB & = CD \phantom{000} (\triangle ABD \equiv \triangle CDB) \end{align} \begin{align} \therefore \text{For parallelogram } ABCD, \phantom{.} AD = BC \text{ and } AB = DC \end{align}

(i)

\begin{align} \angle A + \angle B + \angle C + \angle D & = 360^\circ \phantom{000} (\angle \text{ sum of quadrilateral}) \\ \angle A + \angle B + \angle A + \angle B & = 360^\circ \phantom{000} (\angle A = \angle C \text{ and } \angle B = \angle D) \\ 2\angle A + 2 \angle B & = 360^\circ \\ 2(\angle A + \angle B) & = 360^\circ \\ \angle A + \angle B & = {360^\circ \over 2} \\ \angle A + \angle B & = 180^\circ \text{ (Shown)} \end{align}

(ii)

\begin{align} \angle A + \angle B & = 180^\circ \phantom{000} (\text{From part i}) \\ \\ \therefore AD \phantom{.} & // \phantom{.} BC \phantom{000} (\text{Converse of interior } \angle s) \\ \\ \\ \angle A + \angle B & = 180^\circ \\ \\ \angle A + \angle D & = 180^\circ \phantom{000} (\angle B = \angle D) \\ \\ \therefore AB \phantom{.} & // \phantom{.} DC \phantom{000} (\text{Converse of interior } \angle s) \\ \\ \\ \text{Since } AD \phantom{.} // & \phantom{.} BC \text{ and } AB \phantom{.} // \phantom{.} DC, \phantom{.} ABCD \text{ is a parallelogram} \end{align}

\begin{align} \angle ABC & = 180^\circ - \angle XBA \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line)} \\ & = 180^\circ - \angle YDC \phantom{000} (\text{Given}) \\ & = \angle CDA \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line)} \text{ [A]} \\ \\ \angle BAC & = \angle DCA \phantom{000} \text{ (Given) [A]} \\ \\ AC & \text{ is the common side [S]} \\ \\ \\ \therefore \triangle BAC & \text{ is congruent to } \triangle DCA \text{ (AAS)} \\ \\ \\ \angle BAC & = \angle DAC \phantom{000} (\triangle BAC \equiv \triangle DCA) \\ \\ \therefore AD \phantom{.} & // \phantom{.} BC \phantom{000} (\text{Converse of alternate } \angle s) \\ \\ \\ \angle BAC & = \angle ACD \phantom{000} (\text{Given}) \\ \\ \therefore AB \phantom{.} & // \phantom{.} DC \phantom{000} (\text{Converse of alternate } \angle s) \\ \\ \\ \text{Since } AD \phantom{.} // & \phantom{.} BC \text{ and } AB \phantom{.} // \phantom{.} DC, \phantom{.} ABCD \text{ is a parallelogram} \end{align}

\begin{align} DA \phantom{.} & // \phantom{.} CB \phantom{000} (\text{Opposite sides of parallelogram are parallel}) \\ \\ \therefore DX \phantom{.} & // \phantom{.} YB \phantom{000} (DAX \text{ and } YCB \text{ are straight lines}) \\ \\ \\ \angle ADC & = \angle ABC \phantom{000} (\text{Opposite angles of a parallelogram are equal}) \\ \\ \text{Let } & \angle ABC \text{ be } y^\circ. \\ \\ \angle XAB & = \angle ABC \phantom{000} (DX \phantom{.} // \phantom{.} YB, \text{ alternate } \angle s) \\ & = y^\circ \\ \\ \text{Let } & \angle AXB \text{ be } x^\circ. \\ \\ \angle XBA & = 180^\circ - x^\circ - y^\circ \phantom{000} (\angle \text{ sum of triangle}) \\ & = (180 - x - y)^\circ \\ \\ \\ AX & = CY \phantom{000} \text{(Given) [S]} \\ \\ \angle XAB & = \angle XDC \phantom{000} (AB \phantom{.} // \phantom{.} DC, \text{ Corresponding } \angle s) \\ & = \angle YCD \phantom{000} (DX \phantom{.} // \phantom{.} YB, \text{ Alternate } \angle s) \text{ [A]} \\ \\ AB & = CD \phantom{000} \text{(Opposite sides of parallelogram are equal) [S]} \\ \\ \\ \triangle AXB & \text{ is congruent to } \triangle CYD \text{ (SAS)} \\ \\ \\ \angle AXB & = \angle CYD \phantom{000} (\triangle AXB \equiv \triangle CYD) \\ & = x^\circ \\ \\ \angle XBY + \angle DYB & = (\angle XBA + \angle ABC) + \angle DYB \\ & = \angle XBA + \angle ABC + \angle CYD \phantom{000} (\text{Common } \angle) \\ & = (180 - x - y) + y + x \\ & = 180^\circ \\ \\ \therefore XB \phantom{.} & // \phantom{.} DY \phantom{000} (\text{Converse of interior } \angle s) \\ \\ \\ \text{Since } DX \phantom{.} // & \phantom{.} YB \text{ and } XB \phantom{.} // \phantom{.} DY, \phantom{.} XBYD \text{ is a parallelogram} \end{align}

(i)

\begin{align} AB & = CD \phantom{000} (\text{Opposite sides of parallelogram are equal) [S]} \\ \\ \angle ABE & = \angle CDF \phantom{000} ( AB \phantom{.} // \phantom{.} CD, \text{ Alternate } \angle s ) \text{ [A]} \\ \\ BE & = DF \phantom{000} \text{ (Given) [S]} \\ \\ \\ \therefore \triangle ABE & \text{ is congruent to } \triangle CDF \text{ (SAS)} \end{align}

(ii)

\begin{align} \angle AEF & = 180^\circ - \angle AEB \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line}) \\ & = 180^\circ - \angle CFD \phantom{000} (\triangle ABE \equiv \triangle CDF) \\ & = \angle CFE \\ \\ \angle AEF & = \angle CFE \\ \\ \therefore AE \phantom{.} & // \phantom{.} FC \phantom{000} (\text{Converse of alternate } \angle s) \\ \\ \\\ BC & = DA \phantom{000} (\text{Opposite sides of a parallelogram are equal) [S]} \\ \\ \angle EBC & = \angle FDA \phantom{000} (AD \phantom{.} // \phantom{.} BC, \text{ alternate } \angle s) \text{ [A]} \\ \\ BE & = DF \phantom{000} (\text{Given) [S]} \\ \\ \\ \triangle BEC & \text{ is congruent to } \triangle DFA \\ \\ \\ \angle FEC & = 180^\circ - \angle BEC \phantom{000} (\text{Adjacent } \angle s \text{ on a straight line}) \\ & = 180^\circ - \angle DFA \phantom{000} (\triangle BEC \equiv \triangle DFA) \\ & = \angle AFE \\ \\ \angle FEC & = \angle AFE \\ \\ \therefore AF \phantom{.} & // \phantom{.} EC \phantom{000} (\text{Converse of alternate } \angle s) \\ \\ \\ \text{Since } AE \phantom{.} // & \phantom{.} FC \text{ and } AF \phantom{.} // \phantom{.} EC, \phantom{.} AECF \text{ is a parallelogram} \end{align}

(iii)

\begin{align} \text{Since } \triangle & ABE \equiv \triangle CDF, \\ AE & = CF \\ & = 7 \text{ cm} \end{align}

(i)

\begin{align} AB & = AD \phantom{000} (\text{Adjacent sides of a kite) [S]} \\ \\ BC & = DC \phantom{000} (\text{Adjacent sides of a kite) [S]} \\ \\ AC & \text{ is the common side [S]} \\ \\ \\ \therefore \triangle ABC & \text{ is congruent to } \triangle ADC \text{ (SSS)} \end{align}

(ii)

\begin{align} BC & = DC \phantom{000} (\text{Adjacent sides of a kite) [S]} \\ \\ \angle ACB & = \angle ACD \phantom{000} (\triangle ABC \equiv \triangle ADC) \\ \\ \therefore \angle BCE & = \angle DCE \phantom{000} (\text{Common } \angle) \text{ [A]} \\ \\ CE & \text{ is the common length [S]} \\ \\ \\ \therefore \triangle BCE & \text{ is congruent to } \triangle DCE \text{ (SAS)} \end{align}

(iii)

\begin{align} \angle BEC + \angle DEC & = 180^\circ \phantom{000} \text{(Adjacent } \angle s \text{ on a straight line}) \\ \\ \angle BEC + \angle BEC & = 180^\circ \phantom{000} (\angle BEC = \angle DEC \text{ since } \triangle BCE \equiv \triangle DCE) \\ \\ 2\angle BEC & = 180^\circ \\ \\ \angle BEC & = {180^\circ \over 2} \\ & = 90^\circ \\ \\ \\ \text{Since } \angle BEC = \angle & DEC = 90^\circ, \phantom{.} BD \text{ is perpendicular to } AC \end{align}

(i) Consider triangles ABD and CBD

\begin{align} \text{Since } E \text{ is the mid-} & \text{point of } AB \text{ and } H \text{ is the mid-point of } AD, \\ EH \phantom{.} & // \phantom{.} BD \phantom{000} (\text{Mid-point theorem}) \\ \\ \\ \text{Since } F \text{ is the mid-} & \text{point of } CB \text{ and } G \text{ is the mid-point of } CD, \\ BD \phantom{.} & // \phantom{.} FG \phantom{000} (\text{Mid-point theorem}) \\ \\ \therefore EH \phantom{.} & // \phantom{.} GF \text{ (Shown)} \\ \\ \\ \angle FID & = \angle EIB \phantom{000} \text{ (Vertically opposite }\angle s) \\ & = 90^\circ \\ \\ \angle EFG & = 180^\circ - 90^\circ \phantom{000} \text{ (Interior } \angle s, BD \phantom{0} // \phantom{0} FG) \\ & = 90^\circ \text{ (Shown)} \end{align}

(ii) Note: Draw a line to join points A and C

\begin{align} \angle FEH & = 180^\circ - 90^\circ \phantom{000} \text{ (Interior } \angle s, EH \phantom{0} // \phantom{0} FG) \\ & = 90^\circ \\ \\ \\ \text{Since } E \text{ is the mid-} & \text{point of } BA \text{ and } F \text{ is the mid-point of } BC, \\ EF \phantom{.} & // \phantom{.} AC \phantom{000} (\text{Mid-point theorem}) \\ \\ \text{Since } H \text{ is the mid-} & \text{point of } DA \text{ and } G \text{ is the mid-point of } DC, \\ HG \phantom{.} & // \phantom{.} AC \phantom{000} (\text{Mid-point theorem}) \\ \\ \therefore EF \phantom{.} & // \phantom{.} HG \\ \\ \\ \angle EHG & = 180^\circ - \angle FEH \phantom{000} \text{ (Interior } \angle s, EF \phantom{0} // \phantom{0} HG) \\ & = 180^\circ - 90^\circ \\ & = 90^\circ \\ \\ \angle FGH & = 180^\circ - \angle EFG \phantom{000} \text{ (Interior } \angle s, EF \phantom{0} // \phantom{0} HG) \\ & = 180^\circ - 90^\circ \\ & = 90^\circ \\ \\ \\ \therefore \angle FEH & = \angle EHG = \angle FGH = \angle EFG = 90^\circ \\ \\ \\ \text{Since } EFGH \text{ is a } & \text{quadrilateral with 4 right angles, it is a rectangle} \end{align}

(i)

\begin{align} \angle BFC & = \angle DFA \phantom{000} \text{ (Vertically opposite } \angle s ) \text{ [A]} \\ \\ FC & = FA \phantom{000} \text{ (Given) [S]} \\ \\ \angle FCB & = \angle FAD \phantom{000} \text{ (Alternate } \angle s) \text{ [A]} \\ \\ \\ \triangle BFC & \text{ is congruent to } \triangle DFA \text{ (ASA)} \\ \\ BF & = DF \phantom{000} (\triangle BFC \equiv \triangle DFA) \\ \\ \therefore F \text{ is } & \text{the mid-point of } BD \end{align}

(ii)

\begin{align} \text{Since } E \text{ is the mid-} & \text{point of } AB \text{ and } F \text{ is the mid-point of } AC, \\ \\ EF \phantom{.} // \phantom{.} BC & \text{ and } EF = {1 \over 2}BC \phantom{000} (\text{Mid-point theorem}) \\ \\ \\ \therefore EG \phantom{.} & // \phantom{.} BC \\ \\ \\ \angle FGD & = \angle BCD \phantom{000} \text{ (Corresponding } \angle) \text{ [A]} \\ \\ \angle FDG & = \angle BDC \phantom{000} \text{ (Common } \angle) \text{ [A]} \\ \\ \\ \therefore \triangle FGD & \text{ is similar to } \triangle BCD \text{ (AA Similarity)} \\ \\ \\ \text{Since } \triangle FGD & \text{ is similar to } \triangle BCD, \\ {GD \over CD} & = {FD \over BD} \\ {GD \over CD} & = {FD \over 2FD} \phantom{000} (\text{From part i, } F \text{ is the mid-point of } BD) \\ {GD \over CD} & = {1 \over 2} \\ 2GD & = CD \\ 2GD & = CG + GD \\ GD & = CG \\ \\ \therefore G \text{ is} & \text{ the mid-point of } CD \end{align}

(iii)

\begin{align} FA & = FC \phantom{000} \text{ (Given) [S]} \\ \\ \angle AFB & = \angle CFD \phantom{000} \text{ (Vertically opposite }\angle s) \text{ [A]} \\ \\ FB & = FD \phantom{000} (\text{From part i, } F \text{ is the mid-point of } BD) \text{ [S]} \\ \\ \\ \therefore \triangle AFB & \text{ is congruent to } \triangle CFD \text{ (SAS)} \\ \\ \\ \angle FAB & = \angle FCD \phantom{000} (\triangle AFB \equiv \triangle CFD) \\ \\ \therefore AB \phantom{.} & // \phantom{.} DC \phantom{000} (\text{Converse of alternate } \angle s) \\ \\ \\ \text{Since } AB \phantom{.} // & \phantom{.} DC \text{ and } AD \phantom{.} // \phantom{.} BC, \phantom{.} ABCD \text{ is a parallelogram} \end{align}

(i) Since AB = CD, trapezium ABCD is an isosceles trapezium

\begin{align} BA & = CD \phantom{000} \text{ (Given) [S]} \\ \\ \angle BAD & = \angle CDA \phantom{000} \text{ (Isosceles trapezium) [A]} \\ \\ AD & \text{ is a common length [S]} \\ \\ \\ \therefore \triangle BAD & \text{ is congruent to } \triangle CDA \text{ (SAS)} \end{align}

(ii)

\begin{align} \angle AEB & = \angle DEC \phantom{000} (\text{Vertically opposite } \angle s) \text{ [A]} \\ \\ \angle ABD & = \angle DCA \phantom{000} (\triangle BAD \equiv \triangle CDA) \\ \\ \therefore \angle ABE & = \angle DCE \phantom{000} \text{ [A]} \\ \\ AB & = DC \phantom{000} \text{ [S]} \\ \\ \\ \therefore\triangle AEB & \text{ is congruent to } \triangle DEC \text{ (AAS)} \\ \\ \therefore AE & = DE \phantom{000} (\triangle AEB \equiv \triangle DEC) \text{ (Shown)} \end{align}

(iii)

\begin{align} \angle AED & = \angle CEB \phantom{000} \text{ (Vertically opposite } \angle s) \\ \\ \angle ADE & = \angle CBE \phantom{000} \text{ (Alternate } \angle s) \\ \\ \\ \triangle AED & \text{ is similar to } \triangle CEB \text{ (AA Similarity)} \\ \\ \\ AE & = DE \phantom{000} (\text{From part ii)} \\ \\ \therefore CE & = BE \phantom{000} (\triangle AED \text{ similar to } \triangle CEB \text{ and } AE = DE) \\ \\ \\ \text{Since } BE & = EC, \phantom{.} \triangle BEC \text{ is an isosceles triangle} \end{align}

(i)

\begin{align} BE & = CD \phantom{000} (ECDE \text{ is a rectangle) [S]} \\ \\ \angle AEB & = \angle BED + \angle AED \\ & = \angle CDE + \angle AED \phantom{000} (\angle BED = \angle CDE, \phantom{.} ECDE \text{ is a rectangle}) \\ & = \angle CDE + \angle ADE \phantom{000} (\angle AED = \angle ADE, \text{ isosceles triangle } AED) \\ & = \angle ADC \\ \\ \angle AEB & = \angle ADC \phantom{000} \text{ [A]} \\ \\ AE & = AD \phantom{000} \text{ (Given) [S]} \\ \\ \\ \therefore \triangle ABE & \text{ is congruent to } \triangle ACD \text{ (SAS)} \end{align}

(ii)

\begin{align} \angle BAC & = \angle FAG \phantom{000} (\text{Common } \angle) \text{ [A]} \\ \\ \angle ABC & = \angle AFG \phantom{000} (\text{Corresponding } \angle s, \phantom{.} ED \phantom{.} // \phantom{.} BC \text{ since } BCDE \text{ is a rectangle}) \text{ [A]} \\ \\ \therefore \triangle ABC & \text{ is similar to } \triangle AFG \\ \\ \\ AB & = AC \phantom{000} (\triangle ABE \equiv \triangle ACD) \\ \\ AF & = AG \phantom{000} (\text{Since } \triangle ABC \text{ is similar to } \triangle AFG \text{ and } AB = AC) \text{ [S]} \\ \\ \angle EAF & = \angle EAB \phantom{000} (\text{Common } \angle) \\ & = \angle DAC \phantom{000} (\triangle ABE \equiv \triangle ACD) \\ & = \angle DAG \phantom{000} (\text{Common } \angle) \\ \\ \angle EAF & = \angle DAG \phantom{000} \text{ [A]} \\ \\ AE & = AD \phantom{000} \text{ (Given) [S]} \\ \\ \\ \therefore \triangle AEF & \text{ is congruent to } \triangle ADG \text{ (SAS)} \end{align}
(Consider right angled triangle BEF)
\begin{align} \sin \angle EBF & = {EF \over BF} \\ & = {EF \over 2EF} \phantom{000} (BF = 2EF) \\ & = {1 \over 2} \\ \\ \angle EBF & = \sin^{-1} \left(1 \over 2\right) \\ & = 30^\circ \\ \\ \angle ABE & = \angle EBF \\ & = 30^\circ \\ \\ \angle ACD & = \angle ABE \phantom{000} (\triangle ABE \text{ is congruent to } \triangle ACD) \\ & = 30^\circ \\ \\ \angle ABC & = 90^\circ - \angle ABE \\ & = 90^\circ - 30^\circ \\ & = 60^\circ \\ \\ \angle ACB & = 90^\circ - \angle ACD \\ & = 90^\circ - 30^\circ \\ & = 60^\circ \\ \\ \angle BAC & = 180^\circ - 60^\circ - 60^\circ \phantom{000} (\angle \text{ sum of } \triangle) \\ & = 60^\circ \\ \\ \\ \text{Since } \angle ABC & = \angle ACB = \angle BAC, \phantom{.} \triangle ABC \text{ is equilateral} \end{align}

(i) Consider triangle CAE

\begin{align} \text{Since } X \text{ is the mid-} & \text{point of } AC \text{ and } Y \text{ is the mid-point of } CE, \\ \\ AE \phantom{.} & // \phantom{.} XY \phantom{000} (\text{Mid-point theorem}) \end{align}

(ii) Consider triangle DBF

\begin{align} \text{Since } X \text{ is the mid-} & \text{point of } BD \text{ and } Y \text{ is the mid-point of } DF, \\ \\ XY \phantom{.} & // \phantom{.} BF \phantom{000} (\text{Mid-point theorem}) \\ \\ AE \phantom{.} & // \phantom{.} XY \phantom{000} (\text{From part i}) \\ \\ \therefore AE \phantom{.} & // \phantom{.} BF \end{align}

(iii)

\begin{align} AB \phantom{0} & // \phantom{0} EF \phantom{0} (ABCD \text{ is a parallelogram)} \\ \\ AE \phantom{0} & // \phantom{0} BF \phantom{0} \text{ (From part ii)} \\ \\ \text{Since } AB \phantom{.} // & \phantom{.} EF \text{ and } AE \phantom{.} // \phantom{.} BF, \phantom{.} BAEF \text{ is a parallelogram} \end{align}

\begin{align} AB & = CD \phantom{00000} (\text{Parallelogram } ABCD) [S] \\ \\ \angle EAB & = \angle FCD \phantom{00} (\text{Alternate angles}) [A] \\ \\ AE & = CF \phantom{00000} (\text{Given)} [S] \\ \\ \text{Triangles } & EAB \text{ and } FCD \text{ are congruent (SAS)} \\ \\ \\ AE & = CF \phantom{00000} (\text{Given)} [S] \\ \\ \angle EAD & = \angle FCB \phantom{00} (\text{Alternate angles}) [A] \\ \\ AD & = CB \phantom{00000} (\text{Parallelogram } ABCD) [S] \\ \\ \text{Triangles } & EAD \text{ and } FCB \text{ are congruent (SAS)} \\ \\ \\ EB & = FD \phantom{000000} (\text{Congruent triangles } EAB \text{ and } FCD) \\ \\ ED & = FB \phantom{000000} (\text{Congruent triangles } EAD \text{ and } FCB) \\ \\ \therefore EBFD & \text{ is a parallelogram since opposite sides are equal} \\ \\ \\ HJ & \phantom{.} // \phantom{.} IG \phantom{0000000} (\text{Given)} \\ \\ JG & \phantom{.} // \phantom{.} IH \phantom{0000000} (\text{Parallelogram } EBFD) \\ \\ \therefore HJGI & \text{ is a parallelogram since opposite sides are parallel} \end{align}