Textbook - Triangles Questions in English

(A) $(i)$ All circles have the same shape but not necessarily the same size,therefore they are similar.
$(ii)$ All squares have the same shape (all angles are $90^{\circ}$ and sides are in the same ratio),therefore they are similar.
$(iii)$ All equilateral triangles have angles of $60^{\circ}$ and equal side ratios,therefore they are similar.
$(iv)$ Two polygons are similar if $(a)$ their corresponding angles are equal and $(b)$ their corresponding sides are proportional.

(N/A) Two figures are said to be similar if they have the same shape but not necessarily the same size.
Example $1$: Two equilateral triangles with side lengths $1\, cm$ and $2\, cm$ respectively. Since all equilateral triangles have angles of $60^{\circ}$,they have the same shape.
Example $2$: Two squares with side lengths $1\, cm$ and $2\, cm$ respectively. Since all squares have interior angles of $90^{\circ}$ and proportional sides,they have the same shape.

(N/A) Two figures are said to be non-similar if they do not have the same shape,even if they have the same number of sides.
Example $1$: $A$ trapezium and a square are non-similar figures because their corresponding angles are not equal and their corresponding sides are not in the same ratio.
Example $2$: $A$ triangle and a parallelogram are non-similar figures because they have a different number of sides and different geometric properties.

(N/A) Two polygons are similar if:
$1$. Their corresponding angles are equal.
$2$. Their corresponding sides are in the same ratio (proportional).
In quadrilateral $PQRS$ and $ABCD$:
- The sides are in the ratio $PQ/AB = QR/BC = RS/CD = SP/DA = 1.5/3 = 1/2$. Thus,the sides are proportional.
- However,the angles of $PQRS$ are not $90^{\circ}$,while all angles of $ABCD$ are $90^{\circ}$.
Since the corresponding angles are not equal,the quadrilaterals $PQRS$ and $ABCD$ are not similar.

(N/A) Given: In $\Delta ABC$,$DE || BC$,where $D$ and $E$ are points on $AB$ and $AC$ respectively.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,the other two sides are divided in the same ratio.
Therefore,$\frac{AD}{DB} = \frac{AE}{EC}$.
Taking the reciprocal of both sides:
$\frac{DB}{AD} = \frac{EC}{AE}$.
Adding $1$ to both sides:
$\frac{DB}{AD} + 1 = \frac{EC}{AE} + 1$.
Taking the common denominator:
$\frac{DB + AD}{AD} = \frac{EC + AE}{AE}$.
Since $DB + AD = AB$ and $EC + AE = AC$,we get:
$\frac{AB}{AD} = \frac{AC}{AE}$.
Taking the reciprocal again:
$\frac{AD}{AB} = \frac{AE}{AC}$.
Hence,it is proved.

(N/A) Let us join $AC$ to intersect $EF$ at $G$ (see Figure).
Given: $AB \parallel DC$ and $EF \parallel AB$.
Since lines parallel to the same line are parallel to each other,we have $EF \parallel DC$.
Now,in $\Delta ADC$,since $EG \parallel DC$ (as $EF \parallel DC$),by Basic Proportionality Theorem,we have:
$\frac{AE}{ED} = \frac{AG}{GC} \quad ...(1)$
Similarly,in $\Delta CAB$,since $GF \parallel AB$ (as $EF \parallel AB$),by Basic Proportionality Theorem,we have:
$\frac{CG}{AG} = \frac{CF}{BF}$
Taking the reciprocal of both sides,we get:
$\frac{AG}{GC} = \frac{BF}{FC} \quad ...(2)$
From equations $(1)$ and $(2)$,we get:
$\frac{AE}{ED} = \frac{BF}{FC}$.

(N/A) It is given that $\frac{PS}{SQ} = \frac{PT}{TR}$.
By the Converse of Thales Theorem (Basic Proportionality Theorem),$ST \parallel QR$.
Therefore,$\angle PST = \angle PQR$ (Corresponding angles) $...(1)$
Also,it is given that $\angle PST = \angle PRQ$ $...(2)$
From equations $(1)$ and $(2)$,we get $\angle PQR = \angle PRQ$.
Since the angles opposite to the sides $PQ$ and $PR$ are equal,the sides themselves must be equal.
Therefore,$PQ = PR$.
Hence,$\triangle PQR$ is an isosceles triangle.

(N/A) $(i)$ Let $EC = x \text{ cm}.$
It is given that $DE || BC.$
By using the Basic Proportionality Theorem,we obtain:
$\frac{AD}{DB} = \frac{AE}{EC}$
$\frac{1.5}{3} = \frac{1}{x}$
$x = \frac{3 \times 1}{1.5}$
$x = 2$
$\therefore EC = 2 \text{ cm}.$
$(ii)$ Let $AD = x \text{ cm}.$
It is given that $DE || BC.$
By using the Basic Proportionality Theorem,we obtain:
$\frac{AD}{DB} = \frac{AE}{EC}$
$\frac{x}{7.2} = \frac{1.8}{5.4}$
$x = \frac{1.8 \times 7.2}{5.4}$
$x = 2.4$
$\therefore AD = 2.4 \text{ cm}.$

(N/A) According to the Converse of Thales Theorem (Basic Proportionality Theorem),$EF || QR$ if and only if $\frac{PE}{EQ} = \frac{PF}{FR}$.
Given values are:
$PE = 3.9 \ cm$
$EQ = 3 \ cm$
$PF = 3.6 \ cm$
$FR = 2.4 \ cm$
Calculating the ratios:
$\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$
$\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5$
Since $\frac{PE}{EQ} \neq \frac{PF}{FR}$ $(1.3 \neq 1.5)$,the condition for $EF$ being parallel to $QR$ is not satisfied.
Therefore,$EF$ is not parallel to $QR$.

(N/A) Given:
$PE = 4 \, cm, QE = 4.5 \, cm, PF = 8 \, cm, RF = 9 \, cm$.
According to the Converse of the Basic Proportionality Theorem (Thales Theorem),if a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side.
Calculate the ratios:
$\frac{PE}{EQ} = \frac{4}{4.5} = \frac{40}{45} = \frac{8}{9}$.
$\frac{PF}{FR} = \frac{8}{9}$.
Since $\frac{PE}{EQ} = \frac{PF}{FR}$,by the Converse of the Basic Proportionality Theorem,$EF$ is parallel to $QR$ $(EF || QR)$.

(A) Given: $PQ = 1.28 \, cm, PR = 2.56 \, cm, PE = 0.18 \, cm, PF = 0.36 \, cm$.
According to the converse of the Basic Proportionality Theorem (Thales Theorem),if a line divides two sides of a triangle in the same ratio,then the line is parallel to the third side.
Calculate the ratios:
$\frac{PE}{PQ} = \frac{0.18}{1.28} = \frac{18}{128} = \frac{9}{64}$
$\frac{PF}{PR} = \frac{0.36}{2.56} = \frac{36}{256} = \frac{9}{64}$
Since $\frac{PE}{PQ} = \frac{PF}{PR} = \frac{9}{64}$,the line $EF$ divides the sides $PQ$ and $PR$ in the same ratio.
Therefore,by the converse of the Basic Proportionality Theorem,$EF || QR$.

(N/A) In the given figure,$LM \parallel CB$.
By using the Basic Proportionality Theorem (Thales Theorem) in $\triangle ABC$,we obtain:
$\frac{AM}{AB} = \frac{AL}{AC} \quad ...(i)$
Similarly,in $\triangle ADC$,since $LN \parallel CD$,by using the Basic Proportionality Theorem,we obtain:
$\frac{AN}{AD} = \frac{AL}{AC} \quad ...(ii)$
From equation $(i)$ and equation $(ii)$,we observe that the right-hand sides are equal.
Therefore,we obtain:
$\frac{AM}{AB} = \frac{AN}{AD}$

(N/A) In $\Delta ABC$,$DE \parallel AC$.
By the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{BD}{DA} = \frac{BE}{EC} \quad ...(i)$
In $\Delta BAE$,$DF \parallel AE$.
By the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{BD}{DA} = \frac{BF}{FE} \quad ...(ii)$
From equations $(i)$ and $(ii)$,since the left-hand sides are equal,the right-hand sides must also be equal:
$\frac{BF}{FE} = \frac{BE}{EC}$
Hence,it is proved.

(N/A) In $\Delta POQ$,$DE \parallel OQ$.
Therefore,$\frac{PE}{EQ} = \frac{PD}{DO}$ (Basic Proportionality Theorem) $...(i)$
In $\Delta POR$,$DF \parallel OR$.
Therefore,$\frac{PF}{FR} = \frac{PD}{DO}$ (Basic Proportionality Theorem) $...(ii)$
From $(i)$ and $(ii)$,we obtain:
$\frac{PE}{EQ} = \frac{PF}{FR}$
Therefore,$EF \parallel QR$ (Converse of Basic Proportionality Theorem).

(N/A) In $\Delta POQ$,$AB \parallel PQ$.
Therefore,$\frac{OA}{AP} = \frac{OB}{BQ}$ (Basic Proportionality Theorem) $...(i)$
In $\Delta POR$,$AC \parallel PR$.
Therefore,$\frac{OA}{AP} = \frac{OC}{CR}$ (Basic Proportionality Theorem) $...(ii)$
From $(i)$ and $(ii)$,we obtain:
$\frac{OB}{BQ} = \frac{OC}{CR}$
Therefore,$BC \parallel QR$ (By the converse of the Basic Proportionality Theorem).

(N/A) Consider a triangle $ABC$ in which $P$ is the mid-point of side $AB$ and a line $PQ$ is drawn parallel to $BC$ such that it intersects $AC$ at $Q$.
According to the Basic Proportionality Theorem (Theorem $6.1$),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,the other two sides are divided in the same ratio.
Therefore,we have:
$\frac{AP}{PB} = \frac{AQ}{QC}$
Since $P$ is the mid-point of $AB$,we have $AP = PB$,which implies $\frac{AP}{PB} = 1$.
Substituting this into the equation,we get:
$1 = \frac{AQ}{QC}$
$\Rightarrow AQ = QC$
This proves that $Q$ is the mid-point of $AC$,meaning the line $PQ$ bisects the third side $AC$.

(N/A) Consider a triangle $ABC$ in which $P$ and $Q$ are the mid-points of sides $AB$ and $AC$ respectively.
Given: $AP = PB$ and $AQ = QC$.
To prove: $PQ \parallel BC$.
Proof:
Since $P$ is the mid-point of $AB$,we have $AP = PB$,which implies $\frac{AP}{PB} = 1$.
Since $Q$ is the mid-point of $AC$,we have $AQ = QC$,which implies $\frac{AQ}{QC} = 1$.
From the above two equations,we get $\frac{AP}{PB} = \frac{AQ}{QC}$.
According to the converse of the Basic Proportionality Theorem (Theorem $6.2$),if a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side.
Therefore,$PQ \parallel BC$.

(N/A) Draw a line $EF$ through point $O$,such that $EF \parallel CD$.
In $\triangle ADC$,since $EO \parallel CD$,by using the Basic Proportionality Theorem $(BPT)$,we obtain:
$\frac{AE}{ED} = \frac{AO}{OC} \quad ...(1)$
Since $AB \parallel CD$ and $EF \parallel CD$,it follows that $EF \parallel AB$.
In $\triangle ABD$,since $EO \parallel AB$,by using the Basic Proportionality Theorem,we obtain:
$\frac{AE}{ED} = \frac{BO}{OD} \quad ...(2)$
From equations $(1)$ and $(2)$,we have:
$\frac{AO}{OC} = \frac{BO}{OD}$
Rearranging the terms,we get:
$\frac{AO}{BO} = \frac{OC}{OD}$

(N/A) Let us consider the following figure for the given question.
Draw a line $OE \parallel AB$ such that $E$ lies on $AD$.
In $\triangle ABD$,since $OE \parallel AB$,by using the Basic Proportionality Theorem $(BPT)$,we obtain:
$\frac{AE}{ED} = \frac{BO}{OD}$ $...(1)$
However,it is given that:
$\frac{AO}{BO} = \frac{CO}{DO}$
Rearranging the terms,we get:
$\frac{AO}{CO} = \frac{BO}{DO}$ $...(2)$
From equations $(1)$ and $(2)$,we obtain:
$\frac{AE}{ED} = \frac{AO}{CO}$
In $\triangle ADC$,since $\frac{AE}{ED} = \frac{AO}{OC}$,by the converse of the Basic Proportionality Theorem,we have:
$EO \parallel DC$
Since we constructed $OE \parallel AB$ and we proved $OE \parallel DC$,it follows that:
$AB \parallel DC$
Therefore,$ABCD$ is a trapezium because one pair of opposite sides is parallel.

(N/A) Given: $PQ || RS$.
To prove: $\Delta POQ \sim \Delta SOR$.
Proof:
In $\Delta POQ$ and $\Delta SOR$:
$1$. $\angle P = \angle S$ (Alternate interior angles,as $PQ || RS$ and $PS$ is a transversal).
$2$. $\angle Q = \angle R$ (Alternate interior angles,as $PQ || RS$ and $QR$ is a transversal).
$3$. $\angle POQ = \angle SOR$ (Vertically opposite angles).
Therefore,by the $AAA$ (Angle-Angle-Angle) similarity criterion,$\Delta POQ \sim \Delta SOR$.

$(40^{\circ})$ In $\Delta ABC$ and $\Delta PQR$,
$\frac{AB}{RQ} = \frac{3.8}{7.6} = \frac{1}{2}$,$\frac{BC}{QP} = \frac{6}{12} = \frac{1}{2}$ and $\frac{CA}{PR} = \frac{3\sqrt{3}}{6\sqrt{3}} = \frac{1}{2}$
That is,$\frac{AB}{RQ} = \frac{BC}{QP} = \frac{CA}{PR}$
So,$\Delta ABC \sim \Delta RQP$ ($SSS$ similarity criterion).
Therefore,$\angle C = \angle P$ (Corresponding angles of similar triangles).
In $\Delta ABC$,by the angle sum property:
$\angle C = 180^{\circ} - \angle A - \angle B$
$\angle C = 180^{\circ} - 80^{\circ} - 60^{\circ} = 40^{\circ}$
Since $\angle C = \angle P$,we have $\angle P = 40^{\circ}$.

(N/A) Given: $OA \cdot OB = OC \cdot OD$
Rearranging the terms,we get:
$\frac{OA}{OC} = \frac{OD}{OB} \quad ...(1)$
Also,we have:
$\angle AOD = \angle COB \quad$ (Vertically opposite angles) $...(2)$
From $(1)$ and $(2)$,by the $SAS$ (Side-Angle-Side) similarity criterion,we have:
$\Delta AOD \sim \Delta COB$
Since the triangles are similar,their corresponding angles are equal:
Therefore,$\angle A = \angle C$ and $\angle D = \angle B$.

(N/A) Let $AB$ denote the lamp-post and $CD$ the girl after walking for $4\, seconds$ away from the lamp-post.
From the figure,you can see that $DE$ is the shadow of the girl. Let $DE$ be $x$ meters.
Now,$BD = 1.2\, m/s \times 4\, s = 4.8\, m$.
Note that in $\Delta ABE$ and $\Delta CDE$:
$\angle B = \angle D = 90^{\circ}$ (Each is $90^{\circ}$ because the lamp-post and the girl are standing vertical to the ground).
$\angle E = \angle E$ (Common angle).
So,$\Delta ABE \sim \Delta CDE$ ($AA$ similarity criterion).
Therefore,$\frac{BE}{DE} = \frac{AB}{CD}$.
Given $AB = 3.6\, m$ and $CD = 90\, cm = 0.9\, m$.
$\frac{4.8 + x}{x} = \frac{3.6}{0.9}$.
$\frac{4.8 + x}{x} = 4$.
$4.8 + x = 4x$.
$3x = 4.8$.
$x = 1.6\, m$.
So,the length of the shadow of the girl after walking for $4\, seconds$ is $1.6\, m$.

(A) Given: $\Delta ABC \sim \Delta PQR$.
Since the triangles are similar,their corresponding sides are proportional and corresponding angles are equal:
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} \quad ...(1)$
$\angle A = \angle P, \angle B = \angle Q, \angle C = \angle R \quad ...(2)$
Since $CM$ and $RN$ are medians,$M$ is the midpoint of $AB$ and $N$ is the midpoint of $PQ$. Therefore,$AB = 2AM$ and $PQ = 2PN$.
$(i)$ From $(1)$,$\frac{2AM}{2PN} = \frac{CA}{RP} \implies \frac{AM}{PN} = \frac{CA}{RP}$.
Also,$\angle MAC = \angle NPR$ (from $(2)$).
By $SAS$ similarity criterion,$\Delta AMC \sim \Delta PNR$.
$(ii)$ Since $\Delta AMC \sim \Delta PNR$,their corresponding sides are proportional:
$\frac{CM}{RN} = \frac{CA}{RP}$.
From $(1)$,$\frac{CA}{RP} = \frac{AB}{PQ}$.
Therefore,$\frac{CM}{RN} = \frac{AB}{PQ}$.
$(iii)$ In $\Delta CMB$ and $\Delta RNQ$:
$\frac{CM}{RN} = \frac{BC}{QR}$ (from $(ii)$ and $(1)$).
$\frac{BC}{QR} = \frac{BM}{QN}$ (since $BC = 2BM$ and $QR = 2QN$ from the median property).
Thus,$\frac{CM}{RN} = \frac{BC}{QR} = \frac{BM}{QN}$.
By $SSS$ similarity criterion,$\Delta CMB \sim \Delta RNQ$.

(A) In $\triangle ABC$ and $\triangle PQR$:
$\angle A = 60^{\circ}, \angle B = 80^{\circ}, \angle C = 40^{\circ}$
$\angle P = 60^{\circ}, \angle Q = 80^{\circ}, \angle R = 40^{\circ}$
Since $\angle A = \angle P = 60^{\circ}$,$\angle B = \angle Q = 80^{\circ}$,and $\angle C = \angle R = 40^{\circ}$,the corresponding angles are equal.
Therefore,by the $AAA$ (Angle-Angle-Angle) similarity criterion,$\triangle ABC \sim \triangle PQR$.

(N/A) In $\triangle ABC$ and $\triangle QRP$:
$\frac{AB}{QR} = \frac{2}{4} = \frac{1}{2}$
$\frac{BC}{RP} = \frac{2.5}{5} = \frac{1}{2}$
$\frac{AC}{QP} = \frac{3}{6} = \frac{1}{2}$
Since the ratios of the corresponding sides are equal,by the $SSS$ (Side-Side-Side) similarity criterion,the triangles are similar.
Therefore,$\triangle ABC \sim \triangle QRP$.

(NONE) In $\triangle LMP$ and $\triangle DEF$:
$\frac{LM}{DE} = \frac{2.7}{4} = 0.675$
$\frac{LP}{DF} = \frac{3}{6} = 0.5$
$\frac{MP}{EF} = \frac{2}{5} = 0.4$
Since the ratios of the corresponding sides are not equal (i.e.,$\frac{LM}{DE} \neq \frac{LP}{DF} \neq \frac{MP}{EF}$),the two triangles are not similar.

(N/A) In $\triangle MNL$ and $\triangle QPR$:
Given $\angle M = \angle Q = 70^{\circ}$.
Also,the ratios of the sides including these angles are:
$\frac{MN}{QP} = \frac{2.5}{5} = \frac{1}{2}$
$\frac{ML}{QR} = \frac{5}{10} = \frac{1}{2}$
Since $\frac{MN}{QP} = \frac{ML}{QR} = \frac{1}{2}$ and $\angle M = \angle Q$,by the $SAS$ (Side-Angle-Side) similarity criterion,the two triangles are similar.
Therefore,$\triangle MNL \sim \triangle QPR$.

(NONE) In $\triangle ABC$ and $\triangle DEF$:
Given $\angle A = 80^{\circ}$ and $\angle F = 80^{\circ}$.
Also,the sides including these angles are:
$AB = 2.5$,$AC$ is not given.
$DF = 5$,$EF = 6$.
Checking the ratio of sides:
$\frac{AB}{DF} = \frac{2.5}{5} = \frac{1}{2} = 0.5$
$\frac{BC}{EF} = \frac{3}{6} = \frac{1}{2} = 0.5$
Since the ratio of two sides is equal and the included angle is not the same (the angle $80^{\circ}$ is at $A$ in $\triangle ABC$ and at $F$ in $\triangle DEF$),the triangles are not similar by the $SAS$ criterion.
Therefore,the given triangles are not similar.

(N/A) In $\Delta DEF$,the sum of angles is $180^{\circ}$.
Given $\angle D = 70^{\circ}$ and $\angle E = 80^{\circ}$.
Therefore,$\angle F = 180^{\circ} - (70^{\circ} + 80^{\circ}) = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
In $\Delta PQR$,the sum of angles is $180^{\circ}$.
Given $\angle Q = 80^{\circ}$ and $\angle R = 30^{\circ}$.
Therefore,$\angle P = 180^{\circ} - (80^{\circ} + 30^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
Comparing the two triangles:
$\angle D = \angle P = 70^{\circ}$
$\angle E = \angle Q = 80^{\circ}$
$\angle F = \angle R = 30^{\circ}$
Since all corresponding angles are equal,by the $AAA$ (Angle-Angle-Angle) similarity criterion,the triangles are similar.
Symbolic form: $\Delta DEF \sim \Delta PQR$.

(N/A) Since $DB$ is a straight line,$\angle DOC$ and $\angle BOC$ form a linear pair.
$\angle DOC + \angle BOC = 180^{\circ}$
$\angle DOC + 125^{\circ} = 180^{\circ}$
$\angle DOC = 180^{\circ} - 125^{\circ} = 55^{\circ}$
In $\triangle ODC$,the sum of the angles is $180^{\circ}$:
$\angle DCO + \angle CDO + \angle DOC = 180^{\circ}$
$\angle DCO + 70^{\circ} + 55^{\circ} = 180^{\circ}$
$\angle DCO + 125^{\circ} = 180^{\circ}$
$\angle DCO = 180^{\circ} - 125^{\circ} = 55^{\circ}$
Given that $\Delta ODC \sim \Delta OBA$,the corresponding angles are equal:
$\angle OAB = \angle OCD = \angle DCO$
Therefore,$\angle OAB = 55^{\circ}$.
Thus,$\angle DOC = 55^{\circ}$,$\angle DCO = 55^{\circ}$,and $\angle OAB = 55^{\circ}$.

(N/A) Given: $ABCD$ is a trapezium with $AB \parallel CD$ and diagonals $AC$ and $BD$ intersecting at $O$.
To prove: $\frac{OA}{OC} = \frac{OB}{OD}$.
Proof:
Consider $\triangle OAB$ and $\triangle OCD$.
$1$. $\angle AOB = \angle COD$ (Vertically opposite angles).
$2$. $\angle OAB = \angle OCD$ (Alternate interior angles,since $AB \parallel CD$ and $AC$ is a transversal).
$3$. $\angle OBA = \angle ODC$ (Alternate interior angles,since $AB \parallel CD$ and $BD$ is a transversal).
Therefore,by $AAA$ similarity criterion,$\triangle OAB \sim \triangle OCD$.
Since the triangles are similar,their corresponding sides are proportional:
$\frac{OA}{OC} = \frac{OB}{OD}$.
Hence proved.

(N/A) In $\Delta PQR$,given $\angle PQR = \angle PRQ$.
Since the angles opposite to equal sides are equal,we have $PQ = PR$ $....(i)$
Given,
$\frac{QR}{QS} = \frac{QT}{PR}$
Using $(i)$,we substitute $PR$ with $PQ$ to obtain:
$\frac{QR}{QS} = \frac{QT}{PQ}$ $....(ii)$
Now,in $\Delta PQS$ and $\Delta TQR$:
$\frac{QR}{QS} = \frac{QT}{PQ}$ $[\text{From } (ii)]$
$\angle Q = \angle Q$ $[\text{Common angle}]$
Therefore,by the $SAS$ similarity criterion,$\Delta PQS \sim \Delta TQR$.

(N/A) In $\Delta RPQ$ and $\Delta RST$:
$\angle RTS = \angle RPQ$ (Given)
$\angle R = \angle R$ (Common angle)
Therefore,$\Delta RPQ \sim \Delta RTS$ (By $AA$ similarity criterion).

(N/A) Given: $\Delta ABE \cong \Delta ACD.$
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$ $...(1)$
And,$AE = AD$ $...(2)$
Now,consider $\Delta ADE$ and $\Delta ABC$:
From equations $(1)$ and $(2)$,we have:
$\frac{AD}{AB} = \frac{AE}{AC}$
Also,$\angle DAE = \angle BAC$ (Common angle).
Therefore,by the $SAS$ (Side-Angle-Side) similarity criterion,$\Delta ADE \sim \Delta ABC.$

(N/A) In $\Delta AEP$ and $\Delta CDP$:
$\angle AEP = \angle CDP = 90^{\circ}$ (Since $AD \perp BC$ and $CE \perp AB$)
$\angle APE = \angle CPD$ (Vertically opposite angles)
Therefore,by the $AA$ similarity criterion,
$\Delta AEP \sim \Delta CDP$.

(N/A) In $\Delta ABD$ and $\Delta CBE$:
$\angle ADB = \angle CEB = 90^{\circ}$ (Since $AD \perp BC$ and $CE \perp AB$)
$\angle ABD = \angle CBE$ (Common angle)
Therefore,by using the $AA$ similarity criterion,
$\Delta ABD \sim \Delta CBE$

(N/A) In $\triangle AEP$ and $\triangle ADB$:
$1. \angle AEP = \angle ADB = 90^{\circ}$ (Given that $AD$ and $CE$ are altitudes).
$2. \angle PAE = \angle DAB$ (Common angle).
Therefore,by the $AA$ (Angle-Angle) similarity criterion,we have:
$\triangle AEP \sim \triangle ADB$.

(N/A) In $\Delta PDC$ and $\Delta BEC$:
$\angle PDC = \angle BEC = 90^{\circ}$ (Since $AD \perp BC$ and $CE \perp AB$)
$\angle PCD = \angle BCE$ (Common angle)
Therefore,by using the $AA$ similarity criterion,
$\Delta PDC \sim \Delta BEC$.

(N/A) In $\triangle ABE$ and $\triangle CFB$:
$1$. $\angle A = \angle C$ (Opposite angles of a parallelogram are equal).
$2$. $\angle AEB = \angle CBF$ (Since $AE \parallel BC$ and $BE$ is a transversal,these are alternate interior angles).
Therefore,by the $AA$ (Angle-Angle) similarity criterion,$\triangle ABE \sim \triangle CFB$.

(N/A) In $\Delta ABC$ and $\Delta AMP$:
$\angle ABC = \angle AMP = 90^{\circ}$ (Given)
$\angle A = \angle A$ (Common angle)
Therefore,$\Delta ABC \sim \Delta AMP$ by $AA$ similarity criterion.
Since the triangles are similar,their corresponding sides are proportional.
Thus,$\frac{CA}{PA} = \frac{BC}{MP}$.

(N/A) It is given that $\Delta ABC \sim \Delta FEG$.
Therefore,$\angle A = \angle F, \angle B = \angle E,$ and $\angle ACB = \angle FGE$.
Since $\angle ACB = \angle FGE,$ their bisectors are also equal.
Therefore,$\angle ACD = \angle FGH$ (Angle bisector).
And,$\angle DCB = \angle HGE$ (Angle bisector).
$(i)$ In $\Delta DCA$ and $\Delta HGF$:
$\angle A = \angle F$ (Given)
$\angle ACD = \angle FGH$ (Proved above)
Therefore,$\Delta DCA \sim \Delta HGF$ (By $AA$ similarity criterion).
Since the triangles are similar,the ratio of their corresponding sides is equal:
$\frac{CD}{GH} = \frac{AC}{FG}$.
$(ii)$ In $\Delta DCB$ and $\Delta HGE$:
$\angle DCB = \angle HGE$ (Proved above)
$\angle B = \angle E$ (Given)
Therefore,$\Delta DCB \sim \Delta HGE$ (By $AA$ similarity criterion).
$(iii)$ In $\Delta DCA$ and $\Delta HGF$:
$\angle A = \angle F$ (Given)
$\angle ACD = \angle FGH$ (Proved above)
Therefore,$\Delta DCA \sim \Delta HGF$ (By $AA$ similarity criterion).

(N/A) Given: $ABC$ is an isosceles triangle with $AB = AC$.
Since $AB = AC$,the angles opposite to these sides are equal,so $\angle ABC = \angle ACB$.
Since $E$ is a point on $CB$ produced,$\angle ABC + \angle ABE = 180^{\circ}$ (linear pair).
Also,$\angle ACB + \angle ECF = 180^{\circ}$ is not directly applicable,but we know $\angle ABC = \angle ACB$.
In $\triangle ABD$ and $\triangle ECF$:
$1$. $\angle ADB = \angle EFC = 90^{\circ}$ (Given).
$2$. $\angle ABD = \angle ECF$ is incorrect; rather,$\angle ABC = \angle ACB$. Since $\angle ABC + \angle ABE = 180^{\circ}$ and $\angle ACB + \angle ECF$ is not the relation,we observe that $\angle ABC = \angle ACB$. Since $\angle ACB = \angle ECF$ (vertically opposite is not applicable here),we use $\angle ABC = \angle ACB$. In $\triangle ABC$,$\angle B = \angle C$. Thus,$\angle ABD = 180^{\circ} - \angle ABC$ and $\angle ECF = \angle ACB$. Actually,$\angle B = \angle C$,so $\angle ABD = 180^{\circ} - \angle B$ and $\angle ECF$ is just $\angle C$. Wait,$\angle ABC = \angle ACB$. Therefore,$\angle ABD = 180^{\circ} - \angle ABC = 180^{\circ} - \angle ACB$. This is not equal to $\angle ECF$. Let us re-evaluate: $\angle ABC = \angle ACB$. Since $E$ is on $CB$ produced,$\angle ECF$ is the same as $\angle ACB$. Thus $\angle B = \angle C$. In $\triangle ABD$ and $\triangle ECF$,$\angle ADB = \angle EFC = 90^{\circ}$ and $\angle B = \angle ECF$ is not correct. The correct relation is $\angle ABC = \angle ACB$. Since $\angle ABC = \angle ACB$,and $\angle ECF$ is the exterior angle,$\angle ECF = \angle ABC$. Thus $\angle B = \angle ECF$. By $AA$ similarity,$\Delta ABD \sim \Delta ECF$.

(N/A) Given: In $\Delta ABC$ and $\Delta PQR$,$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$.
Since $AD$ and $PM$ are medians,$D$ and $M$ are midpoints of $BC$ and $QR$ respectively.
Therefore,$BD = \frac{BC}{2}$ and $QM = \frac{QR}{2}$.
Substituting these in the given ratio:
$\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM} \Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$.
In $\Delta ABD$ and $\Delta PQM$:
$\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$ (Proved above).
Therefore,$\Delta ABD \sim \Delta PQM$ by $SSS$ similarity criterion.
This implies $\angle B = \angle Q$ (Corresponding angles of similar triangles).
Now,in $\Delta ABC$ and $\Delta PQR$:
$1$. $\frac{AB}{PQ} = \frac{BC}{QR}$ (Given)
$2$. $\angle B = \angle Q$ (Proved above)
Therefore,$\Delta ABC \sim \Delta PQR$ by $SAS$ similarity criterion.

(N/A) In $\triangle ADC$ and $\triangle BAC$:
$\angle ADC = \angle BAC$ (Given)
$\angle ACD = \angle BCA$ (Common angle)
Therefore,$\triangle ADC \sim \triangle BAC$ (By $AA$ similarity criterion).
We know that the corresponding sides of similar triangles are proportional.
Therefore,$\frac{CA}{CB} = \frac{CD}{CA}$.
Cross-multiplying gives $CA^2 = CB \cdot CD$.

(A) Given that,
$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$
Let us extend $AD$ and $PM$ up to points $E$ and $L$ respectively,such that $AD = DE$ and $PM = ML$. Then,join $B$ to $E$,$C$ to $E$,$Q$ to $L$,and $R$ to $L$.
We know that medians divide opposite sides. Therefore,$BD = DC$ and $QM = MR$.
Also,$AD = DE$ (by construction) and $PM = ML$ (by construction).
In quadrilateral $ABEC$,diagonals $AE$ and $BC$ bisect each other at point $D$. Therefore,quadrilateral $ABEC$ is a parallelogram.
$\therefore AC = BE$ and $AB = EC$ (opposite sides of a parallelogram are equal).
Similarly,we can prove that quadrilateral $PQLR$ is a parallelogram and $PR = QL$,$PQ = LR$.
It was given that $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$.
$\Rightarrow \frac{AB}{PQ} = \frac{BE}{QL} = \frac{2AD}{2PM}$
$\Rightarrow \frac{AB}{PQ} = \frac{BE}{QL} = \frac{AE}{PL}$
$\therefore \Delta ABE \sim \Delta PQL$ (by $SSS$ similarity criterion).
We know that corresponding angles of similar triangles are equal.
$\therefore \angle BAE = \angle QPL \dots(1)$
Similarly,it can be proved that $\Delta AEC \sim \Delta PLR$ and $\angle CAE = \angle RPL \dots(2)$
Adding equation $(1)$ and $(2)$,we obtain:
$\angle BAE + \angle CAE = \angle QPL + \angle RPL$
$\Rightarrow \angle CAB = \angle RPQ \dots(3)$
In $\Delta ABC$ and $\Delta PQR$:
$\frac{AB}{PQ} = \frac{AC}{PR}$ (given)
$\angle CAB = \angle RPQ$ (using equation $(3)$)
$\therefore \Delta ABC \sim \Delta PQR$ (by $SAS$ similarity criterion).

(C) Let $AB$ be the height of the tower and $CD$ be the height of the pole.
Let $BE$ be the shadow of the tower and $DF$ be the shadow of the pole.
Since the sun's rays fall at the same angle at the same time,the triangles formed by the objects and their shadows are similar.
Therefore,$\triangle ABE \sim \triangle CDF$ by $AA$ similarity criterion.
This implies that the ratios of their corresponding sides are equal:
$\frac{AB}{CD} = \frac{BE}{DF}$
Given $CD = 6\, m$,$DF = 4\, m$,and $BE = 28\, m$.
Substituting the values:
$\frac{AB}{6} = \frac{28}{4}$
$\frac{AB}{6} = 7$
$AB = 7 \times 6 = 42\, m$.
Thus,the height of the tower is $42\, m$.

(N/A) It is given that $\Delta ABC \sim \Delta PQR.$
We know that the corresponding sides of similar triangles are in proportion.
$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR} \dots(1)$
Also,$\angle A = \angle P, \angle B = \angle Q, \angle C = \angle R \dots(2)$
Since $AD$ and $PM$ are medians,they divide their opposite sides into two equal parts.
$BD = \frac{BC}{2}$ and $QM = \frac{QR}{2} \dots(3)$
From equations $(1)$ and $(3),$ we obtain
$\frac{AB}{PQ} = \frac{BC/2}{QR/2} = \frac{BD}{QM} \dots(4)$
In $\Delta ABD$ and $\Delta PQM,$
$\angle B = \angle Q$ [Using equation $(2)$]
$\frac{AB}{PQ} = \frac{BD}{QM}$ [Using equation $(4)$]
$\therefore \Delta ABD \sim \Delta PQM$ (By $SAS$ similarity criterion)
$\Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$

(D) We have $XY \parallel AC$ (Given).
$\angle BXY = \angle A$ and $\angle BYX = \angle C$ (Corresponding angles).
Therefore,$\Delta ABC \sim \Delta XBY$ ($AA$ similarity criterion).
$\frac{\operatorname{ar}(ABC)}{\operatorname{ar}(XBY)} = \left(\frac{AB}{XB}\right)^2$ (Theorem regarding ratio of areas of similar triangles).
Since $XY$ divides the triangle into two parts of equal areas,$\operatorname{ar}(ABC) = 2 \operatorname{ar}(XBY)$.
$\frac{\operatorname{ar}(ABC)}{\operatorname{ar}(XBY)} = \frac{2}{1}$.
Therefore,$\left(\frac{AB}{XB}\right)^2 = \frac{2}{1}$,which implies $\frac{AB}{XB} = \frac{\sqrt{2}}{1}$.
Taking the reciprocal,$\frac{XB}{AB} = \frac{1}{\sqrt{2}}$.
Now,$\frac{AX}{AB} = \frac{AB - XB}{AB} = 1 - \frac{XB}{AB} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$.
Rationalizing the denominator,$\frac{AX}{AB} = \frac{(\sqrt{2} - 1) \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{2 - \sqrt{2}}{2}$.

(B) It is given that $\Delta ABC \sim \Delta DEF$.
According to the theorem of areas of similar triangles,the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta DEF)} = \left(\frac{BC}{EF}\right)^2$
Given that $\text{ar}(\Delta ABC) = 64 \text{ cm}^2$,$\text{ar}(\Delta DEF) = 121 \text{ cm}^2$,and $EF = 15.4 \text{ cm}$.
$\frac{64}{121} = \left(\frac{BC}{15.4}\right)^2$
Taking the square root on both sides:
$\sqrt{\frac{64}{121}} = \frac{BC}{15.4}$
$\frac{8}{11} = \frac{BC}{15.4}$
$BC = \frac{8 \times 15.4}{11}$
$BC = 8 \times 1.4 = 11.2 \text{ cm}$.