Criteria for Similarity of Triangles Questions in English

(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 $\angle D = \angle B$.

(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,$\triangle ABE \sim \triangle CFB$.

(C) In $\triangle ADB$ and $\triangle ADC$:
$\angle ADB = \angle ADC = 90^{\circ}$ (since $AD \perp BC$)
$\angle BAD = \angle ACD$ (both are complementary to $\angle CAD$)
$\angle ABD = \angle CAD$ (both are complementary to $\angle BAD$)
Therefore,by $AA$ similarity criterion,$\triangle ADB \sim \triangle CDA$.
From the property of similar triangles,the ratio of corresponding sides is equal:
$\frac{AD}{CD} = \frac{BD}{AD}$
Cross-multiplying gives:
$AD^{2} = BD \cdot CD$

(B) Given,$\triangle ABC \sim \triangle EDF$.
Therefore,the ratios of corresponding sides are equal:
$\frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF}$.
From $\frac{AB}{ED} = \frac{BC}{DF}$,we get $AB \cdot DF = ED \cdot BC$,which is $BC \cdot DE = AB \cdot DF$. Thus,option $(d)$ is true.
From $\frac{BC}{DF} = \frac{AC}{EF}$,we get $BC \cdot EF = AC \cdot DF$. This does not match any option directly,but let's check the others.
From $\frac{AB}{ED} = \frac{AC}{EF}$,we get $AB \cdot EF = AC \cdot ED$,which is $AB \cdot EF = AC \cdot DE$. Thus,option $(a)$ is true.
Comparing the given options with the derived ratios:
Option $(a)$: $AB \cdot EF = AC \cdot DE$ (True)
Option $(d)$: $BC \cdot DE = AB \cdot DF$ (True)
Option $(b)$: $BC \cdot DE = AB \cdot EF$ (This is not derived from the similarity $\triangle ABC \sim \triangle EDF$)
Option $(c)$: $BC \cdot EF = AC \cdot FD$ (This is not derived from the similarity $\triangle ABC \sim \triangle EDF$)
Re-evaluating the question: The question asks which is $NOT$ true. Based on the similarity $\triangle ABC \sim \triangle EDF$,the correct ratios are $\frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF}$.
From this,$BC \cdot EF = AC \cdot DF$ is true. Option $(c)$ says $BC \cdot EF = AC \cdot FD$. Since $DF = FD$,option $(c)$ is also true.
Therefore,option $(b)$ is the one that is not true.

(B) Given,in two triangles $ABC$ and $PQR$,$\frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ}$.
This ratio shows that the sides of $\triangle ABC$ are proportional to the corresponding sides of $\triangle PQR$.
Specifically,the side $AB$ corresponds to $QR$,$BC$ corresponds to $PR$,and $CA$ corresponds to $PQ$.
By the $SSS$ (Side-Side-Side) similarity criterion,if the sides of two triangles are in the same ratio,then the triangles are similar,and their corresponding angles are equal.
Matching the vertices based on the proportionality:
$A$ corresponds to $Q$,$B$ corresponds to $R$,and $C$ corresponds to $P$.
Therefore,$\triangle ABC \sim \triangle QRP$,or equivalently,$\triangle CAB \sim \triangle PQR$.

(D) Given,in $\triangle DEF$ and $\triangle PQR$,$\angle D = \angle Q$ and $\angle E = \angle R$.
By the $AA$ similarity criterion,$\triangle DEF \sim \triangle QRP$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{DE}{QR} = \frac{EF}{RP} = \frac{DF}{QP}$.
Now,let's check the given options:
Option $A$: $\frac{EF}{PR} = \frac{DF}{PQ}$ is equivalent to $\frac{EF}{RP} = \frac{DF}{QP}$,which is true.
Option $B$: $\frac{EF}{RP} = \frac{DE}{QR}$ is true.
Option $C$: $\frac{DE}{QR} = \frac{DF}{PQ}$ is equivalent to $\frac{DE}{QR} = \frac{DF}{QP}$,which is true.
Option $D$: $\frac{DE}{PQ} = \frac{EF}{RP}$. From the similarity ratio,$\frac{DE}{QR} = \frac{EF}{RP}$. Thus,$\frac{DE}{PQ} = \frac{EF}{RP}$ is not necessarily true because $PQ \neq QR$ in general.
Therefore,option $D$ is not true.

(A) In $\triangle ABC$ and $\triangle DEF$,we are given $\angle B = \angle E$ and $\angle F = \angle C$.
By the $AA$ (Angle-Angle) similarity criterion,since two angles of $\triangle ABC$ are equal to two corresponding angles of $\triangle DEF$,the triangles are similar,i.e.,$\triangle ABC \sim \triangle DEF$.
For two triangles to be congruent,their corresponding sides must be equal. Here,it is given that $AB = 3 DE$,which implies $AB \neq DE$.
Since the corresponding sides are not equal,the triangles are not congruent.
Therefore,the triangles are similar but not congruent.

(B) Given that $\triangle ABC \sim \triangle PQR$ and $\frac{BC}{QR} = \frac{1}{3}.$
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\operatorname{ar}(\triangle PRQ)}{\operatorname{ar}(\triangle BCA)} = \frac{(QR)^2}{(BC)^2} = \left(\frac{QR}{BC}\right)^2.$
Since $\frac{BC}{QR} = \frac{1}{3},$ it follows that $\frac{QR}{BC} = \frac{3}{1} = 3.$
Thus,$\frac{\operatorname{ar}(\triangle PRQ)}{\operatorname{ar}(\triangle BCA)} = (3)^2 = 9.$

(D) For two triangles to be similar by the $SAS$ (Side-Angle-Side) similarity criterion,the ratio of two corresponding sides must be equal,and the included angle between these sides must be equal.
Given the ratio $\frac{AB}{DE} = \frac{BC}{FD}$,the sides $AB$ and $BC$ form the angle $\angle B$ in $\triangle ABC$. Similarly,the sides $DE$ and $FD$ form the angle $\angle D$ in $\triangle DEF$.
Therefore,for $\triangle ABC \sim \triangle EDF$,the included angles must be equal,i.e.,$\angle B = \angle D$.

(B) No,it is not true.
Given that $\triangle FED \sim \triangle STU$,the corresponding vertices are $F \leftrightarrow S$,$E \leftrightarrow T$,and $D \leftrightarrow U$.
According to the property of similar triangles,the ratios of their corresponding sides must be equal.
Therefore,the correct ratios are $\frac{FE}{ST} = \frac{ED}{TU} = \frac{FD}{SU}$.
Comparing this with the given expression $\frac{DE}{ST} = \frac{EF}{TU}$,we can see that it does not match the required correspondence of sides.

(B) To determine if the triangle is a right triangle,we check if it satisfies the Pythagorean theorem,which states that for a right triangle,the square of the longest side must be equal to the sum of the squares of the other two sides.
Let the sides be $a = 25 \, cm$,$b = 5 \, cm$,and $c = 24 \, cm$.
The longest side is $a = 25 \, cm$.
Calculate the square of the longest side: $a^2 = (25)^2 = 625 \, cm^2$.
Calculate the sum of the squares of the other two sides: $b^2 + c^2 = (5)^2 + (24)^2 = 25 + 576 = 601 \, cm^2$.
Since $a^2 \neq b^2 + c^2$ $(625 \neq 601)$,the triangle does not satisfy the Pythagorean theorem.
Therefore,the triangle with sides $25 \, cm$,$5 \, cm$,and $24 \, cm$ is not a right triangle.

(B) The statement is false.
We know that if two triangles are similar,their corresponding angles are equal in the order of their vertices.
Given $\triangle DEF \sim \triangle RPQ$,the correspondence is:
$D \leftrightarrow R$
$E \leftrightarrow P$
$F \leftrightarrow Q$
Therefore,the correct equalities are $\angle D = \angle R$,$\angle E = \angle P$,and $\angle F = \angle Q$.
Since $\angle F = \angle Q$ and not $\angle P$,the statement $\angle F = \angle P$ is incorrect.

(A) Yes,$\triangle PBC \sim \triangle PDE$.
In $\triangle PBC$ and $\triangle PDE$:
$\angle BPC = \angle EPD$ [Vertically opposite angles]
Now,calculating the ratios of the sides including the angles:
$\frac{PB}{PD} = \frac{5 \text{ cm}}{10 \text{ cm}} = \frac{1}{2}$ ......$(i)$
$\frac{PC}{PE} = \frac{6 \text{ cm}}{12 \text{ cm}} = \frac{1}{2}$ ......$(ii)$
From equations $(i)$ and $(ii)$:
$\frac{PB}{PD} = \frac{PC}{PE}$
Since one angle of $\triangle PBC$ is equal to one angle of $\triangle PDE$ and the sides including these angles are proportional,by the $SAS$ (Side-Angle-Side) similarity criterion,the triangles are similar.
Hence,$\triangle PBC \sim \triangle PDE$.

(A) Yes,the two triangles are similar.
Let the sides of the first triangle be $a_1, b_1, c_1$ and its perimeter be $P_1 = a_1 + b_1 + c_1$.
Let the sides of the second triangle be $a_2, b_2, c_2$ and its perimeter be $P_2 = a_2 + b_2 + c_2$.
Given that $a_1 = 3a_2$,$b_1 = 3b_2$,and $P_1 = 3P_2$.
Since $P_1 = a_1 + b_1 + c_1$ and $P_2 = a_2 + b_2 + c_2$,we have $3P_2 = 3a_2 + 3b_2 + c_1$.
Substituting $P_2 = a_2 + b_2 + c_2$,we get $3(a_2 + b_2 + c_2) = 3a_2 + 3b_2 + c_1$,which simplifies to $3c_2 = c_1$.
Since all three corresponding sides are proportional $(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = 3)$,by the $SSS$ similarity criterion,the two triangles are similar.

(A) Yes,the two triangles will be similar.
Let the two right-angled triangles be $\triangle ABC$ and $\triangle PQR$,where $\angle B = \angle Q = 90^{\circ}$.
Given that one acute angle of the first triangle is equal to an acute angle of the second triangle,let $\angle A = \angle P$.
In $\triangle ABC$ and $\triangle PQR$:
$1$. $\angle B = \angle Q = 90^{\circ}$ (Given)
$2$. $\angle A = \angle P$ (Given)
By the $AA$ (Angle-Angle) similarity criterion,since two angles of one triangle are equal to two angles of the other triangle,the triangles are similar $(\triangle ABC \sim \triangle PQR)$.
This is also a specific case of the $AAA$ similarity criterion,as the third angles must also be equal due to the angle sum property of a triangle.

(B) The statement is False.
According to the theorem on the areas of similar triangles,the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding altitudes.
$\frac{\text{Area}_1}{\text{Area}_2} = \left( \frac{\text{Altitude}_1}{\text{Altitude}_2} \right)^2$
Given that the ratio of the altitudes is $\frac{3}{5}$,we have:
$\frac{\text{Area}_1}{\text{Area}_2} = \left( \frac{3}{5} \right)^2 = \frac{9}{25}$
Since $\frac{9}{25} \neq \frac{6}{5}$,the given statement is incorrect.

(N/A) Yes,it is true.
In $\triangle ADE$ and $\triangle ACB$:
$1$. $\angle A = \angle A$ (Common angle in both triangles).
$2$. $\angle D = \angle C$ (Given in the problem).
Therefore,by the $AA$ similarity criterion,$\triangle ADE \sim \triangle ACB$.

(B) The statement is $False$.
According to the $SAS$ (Side-Angle-Side) similarity criterion,two triangles are similar if one angle of one triangle is equal to one angle of the other triangle and the sides including these angles are proportional.
In the given statement,the two sides are proportional,but they are not necessarily the sides that include the equal angle. Therefore,the condition for $SAS$ similarity is not satisfied,and the triangles are not necessarily similar.

(N/A) Given: In $\triangle PQR$,$PR^{2} - PQ^{2} = QR^{2}$ and $QM \perp PR$.
To prove: $QM^{2} = PM \times MR$.
Proof: Since $PR^{2} - PQ^{2} = QR^{2}$,we have $PR^{2} = PQ^{2} + QR^{2}$.
By the converse of the Pythagoras theorem,$\triangle PQR$ is a right-angled triangle at $Q$.
In $\triangle QMR$ and $\triangle PMQ$:
$\angle M = \angle M = 90^{\circ}$ (given $QM \perp PR$ and $\angle PQR = 90^{\circ}$ implies $\angle MQR + \angle MQP = 90^{\circ}$).
Since $\angle P + \angle R = 90^{\circ}$ and $\angle MQR + \angle R = 90^{\circ}$,we have $\angle MQR = \angle P$.
Similarly,$\angle MQP = \angle R$.
Thus,$\triangle QMR \sim \triangle PMQ$ by $AA$ similarity criterion.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{QM}{PM} = \frac{MR}{QM}$.
Therefore,$QM^{2} = PM \times MR$. Hence proved.

(D) Given $PQRS$ is a trapezium in which $PQ \parallel RS$ and $PQ = 3 RS$.
$\frac{PQ}{RS} = \frac{3}{1}$ ......$(i)$
In $\triangle POQ$ and $\triangle ROS$:
$\angle SOR = \angle QOP$ [Vertically opposite angles]
$\angle OSR = \angle OPQ$ [Alternate interior angles,since $PQ \parallel RS$]
$\therefore \triangle POQ \sim \triangle ROS$ [By $AA$ similarity criterion]
By the property of the area of similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides:
$\frac{\text{ar}(\triangle POQ)}{\text{ar}(\triangle ROS)} = \left(\frac{PQ}{RS}\right)^2 = \left(\frac{3}{1}\right)^2 = \frac{9}{1}$
Hence,the required ratio is $9: 1$.

(N/A) Given: $AC$ and $PQ$ intersect each other at the point $O$ and $AB \parallel DC$.
To prove: $OA \cdot CQ = OC \cdot AP$.
Proof:
In $\triangle AOP$ and $\triangle COQ$:
$1$. $\angle AOP = \angle COQ$ (Vertically opposite angles).
$2$. $\angle OAP = \angle OCQ$ (Alternate interior angles,since $AB \parallel DC$ and $AC$ is a transversal).
$3$. $\angle OPA = \angle OQC$ (Alternate interior angles,since $AB \parallel DC$ and $PQ$ is a transversal).
Therefore,$\triangle AOP \sim \triangle COQ$ (by $AAA$ similarity criterion).
Since the triangles are similar,their corresponding sides are proportional:
$\frac{OA}{OC} = \frac{AP}{CQ}$
By cross-multiplying,we get:
$OA \cdot CQ = OC \cdot AP$.
Hence proved.

(C) Given that $\triangle ABC \sim \triangle DEF$.
Since the triangles are similar,the ratio of their corresponding sides is equal:
$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$.
Substituting the given values:
$\frac{4}{6} = \frac{BC}{9} = \frac{AC}{12}$.
Now,solving for $BC$:
$\frac{4}{6} = \frac{BC}{9} \implies BC = \frac{4 \times 9}{6} = 6 \, cm$.
Solving for $AC$:
$\frac{4}{6} = \frac{AC}{12} \implies AC = \frac{4 \times 12}{6} = 8 \, cm$.
The perimeter of $\triangle ABC = AB + BC + AC$.
Perimeter $= 4 + 6 + 8 = 18 \, cm$.

(D) Given,$DE \parallel BC,$ $DE = 6 \, cm$ and $BC = 12 \, cm.$
In $\triangle ABC$ and $\triangle ADE,$
$\angle ABC = \angle ADE$ [Corresponding angles]
$\angle ACB = \angle AED$ [Corresponding angles]
$\angle A = \angle A$ [Common angle]
Therefore,$\triangle ADE \sim \triangle ABC$ [By $AAA$ similarity criterion].
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\frac{\operatorname{ar}(ADE)}{\operatorname{ar}(ABC)} = \frac{(DE)^2}{(BC)^2}$
$= \frac{(6)^2}{(12)^2} = \left(\frac{6}{12}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}.$
Let $\operatorname{ar}(ADE) = k,$ then $\operatorname{ar}(ABC) = 4k.$
Now,$\operatorname{ar}(DECB) = \operatorname{ar}(ABC) - \operatorname{ar}(ADE) = 4k - k = 3k.$
Therefore,the required ratio is $\operatorname{ar}(ADE) : \operatorname{ar}(DECB) = k : 3k = 1 : 3.$

(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,$\frac{AX}{AB} = \frac{(\sqrt{2} - 1) \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{2 - \sqrt{2}}{2}$.