Criteria for Congruence of Triangles Questions in English

$(i)$ Consider $\Delta AOB$ and $\Delta DOC$.
$\angle OAB = \angle ODC$ (Alternate interior angles as $AB \parallel CD$ and $AD$ is the transversal)
$\angle AOB = \angle DOC$ (Vertically opposite angles)
$OA = OD$ (Given,as $O$ is the mid-point of $AD$)
Therefore,by $ASA$ congruence rule,$\Delta AOB \cong \Delta DOC$.
$(ii)$ Since $\Delta AOB \cong \Delta DOC$,then by $CPCT$ (Corresponding Parts of Congruent Triangles),
$OB = OC$
Thus,$O$ is the mid-point of $BC$.

(N/A) $(i)$ In quadrilateral $ABCD$,we have $AD = BC$ and $\angle DAB = \angle CBA$.
Consider $\Delta ABD$ and $\Delta BAC$:
$AD = BC$ (Given)
$AB = BA$ (Common side)
$\angle DAB = \angle CBA$ (Given)
Therefore,by $SAS$ (Side-Angle-Side) congruence criterion,we have $\Delta ABD \cong \Delta BAC$.
$(ii)$ Since $\Delta ABD \cong \Delta BAC$,their corresponding parts are equal $(CPCT)$.
Therefore,$BD = AC$.
$(iii)$ Since $\Delta ABD \cong \Delta BAC$,their corresponding parts are equal $(CPCT)$.
Therefore,$\angle ABD = \angle BAC$.

(N/A) Given: $l \parallel m$ and $p \parallel q$.
To prove: $\Delta ABC \cong \Delta CDA$.
Proof:
Since $l \parallel m$ and $AC$ is a transversal,
$\therefore \angle BAC = \angle DCA$ [Alternate interior angles]
Also,since $p \parallel q$ and $AC$ is a transversal,
$\therefore \angle BCA = \angle DAC$ [Alternate interior angles]
Now,in $\Delta ABC$ and $\Delta CDA$:
$\angle BAC = \angle DCA$ [Proved above]
$\angle BCA = \angle DAC$ [Proved above]
$AC = CA$ [Common side]
Therefore,by the $ASA$ (Angle-Side-Angle) congruence criterion,$\Delta ABC \cong \Delta CDA$.

(N/A) In $\Delta ABD$ and $\Delta ACD$:
$\angle BAD = \angle CAD$ (Given,as $AD$ is the bisector of $\angle A$)
$AD = AD$ (Common side)
$\angle ADB = \angle ADC = 90^\circ$ (Given,as $AD \perp BC$)
Therefore,by the $ASA$ (Angle-Side-Angle) congruence rule,$\Delta ABD \cong \Delta ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AB = AC$.
Since two sides of $\Delta ABC$ are equal,$\Delta ABC$ is an isosceles triangle.

(N/A) In $\Delta ABF$ and $\Delta ACE$:
$AB = AC$ (Given)
$\angle A = \angle A$ (Common angle)
$AF = AE$ (Since $F$ and $E$ are mid-points of equal sides $AC$ and $AB$ respectively,$AF = \frac{1}{2} AC$ and $AE = \frac{1}{2} AB$. Since $AB = AC$,it follows that $AF = AE$.)
Therefore,$\Delta ABF \cong \Delta ACE$ by the $SAS$ congruence rule.
Since the triangles are congruent,their corresponding parts are equal.
Thus,$BF = CE$ (by $CPCT$).

(N/A) In $\Delta ABD$ and $\Delta ACE$:
$AB = AC$ (Given) ........... $(1)$
$\angle B = \angle C$ (Angles opposite to equal sides) ........... $(2)$
Also,$BE = CD$
Subtracting $DE$ from both sides:
$BE - DE = CD - DE$
$BD = CE$ ........... $(3)$
Therefore,$\Delta ABD \cong \Delta ACE$ (Using $(1)$,$(2)$,$(3)$ and $SAS$ congruence rule).
This gives $AD = AE$ $(CPCT)$.

(N/A) Given: $AD$ is the perpendicular bisector of $BC$,which implies $BD = CD$ and $\angle ADB = \angle ADC = 90^o$.
In $\Delta ABD$ and $\Delta ACD$:
$AD = AD$ (Common side)
$\angle ADB = \angle ADC = 90^o$ (Given)
$BD = CD$ (Since $AD$ is the bisector of $BC$)
Therefore,$\Delta ABD \cong \Delta ACD$ by the $SAS$ congruence criterion.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AB = AC$.
Since two sides of $\Delta ABC$ are equal,it is an isosceles triangle.

(N/A) Given: $\Delta ABC$ is an isosceles triangle with $AB = AC$.
Altitudes $BE \perp AC$ and $CF \perp AB$.
To prove: $BE = CF$.
Proof:
In $\Delta ABE$ and $\Delta ACF$:
$1$. $\angle A = \angle A$ (Common angle)
$2$. $AB = AC$ (Given)
$3$. $\angle AEB = \angle AFC = 90^\circ$ (Since $BE$ and $CF$ are altitudes)
Therefore,by $AAS$ congruence criterion,$\Delta ABE \cong \Delta ACF$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$BE = CF$.

(N/A) Given: $PA = PB$ and $QA = QB$. We need to show that $PQ \perp AB$ and $PQ$ bisects $AB$. Let $PQ$ intersect $AB$ at $C$.
Consider $\Delta PAQ$ and $\Delta PBQ$:
$AP = BP$ (Given)
$AQ = BQ$ (Given)
$PQ = PQ$ (Common side)
Therefore,$\Delta PAQ \cong \Delta PBQ$ by $SSS$ congruence rule.
This implies $\angle APQ = \angle BPQ$ by $CPCT$.
Now,consider $\Delta PAC$ and $\Delta PBC$:
$AP = BP$ (Given)
$\angle APC = \angle BPC$ (Since $\angle APQ = \angle BPQ$)
$PC = PC$ (Common side)
Therefore,$\Delta PAC \cong \Delta PBC$ by $SAS$ congruence rule.
This implies $AC = BC$ and $\angle ACP = \angle BCP$ by $CPCT$.
Since $\angle ACP + \angle BCP = 180^o$ (Linear pair),
$2 \angle ACP = 180^o \implies \angle ACP = 90^o$.
Since $AC = BC$ and $\angle ACP = 90^o$,$PQ$ is the perpendicular bisector of $AB$.

(N/A) Given: Lines $l$ and $m$ intersect at point $A$. $P$ is a point such that $PB \perp l$ and $PC \perp m$,with $PB = PC$.
To prove: Line $AP$ bisects the angle between lines $l$ and $m$,i.e.,$\angle PAB = \angle PAC$.
Proof: Consider $\Delta PAB$ and $\Delta PAC$.
In these two triangles:
$1$. $PB = PC$ (Given,as $P$ is equidistant from the lines)
$2$. $\angle PBA = \angle PCA = 90^o$ (Given,as $PB \perp l$ and $PC \perp m$)
$3$. $PA = PA$ (Common side)
Therefore,by the $RHS$ congruence rule,$\Delta PAB \cong \Delta PAC$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$\angle PAB = \angle PAC$.
This shows that the line $AP$ bisects the angle between lines $l$ and $m$.

(N/A) $(i)$ In $\Delta ABD$ and $\Delta ACD$,we have:
$AB = AC$ [Given]
$\angle ADB = \angle ADC = 90^\circ$ [Since $AD$ is an altitude]
$AD = AD$ [Common side]
Therefore,by $RHS$ congruence rule,$\Delta ABD \cong \Delta ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$BD = CD$,which means $AD$ bisects $BC$.
$(ii)$ Since $\Delta ABD \cong \Delta ACD$,their corresponding angles are equal $(CPCT)$.
Therefore,$\angle BAD = \angle CAD$,which means $AD$ bisects $\angle A$.

(N/A) In $\Delta ABC$,$AM$ is a median [Given].
$\therefore BM = \frac{1}{2} BC$ ........... $(1)$
In $\Delta PQR$,$PN$ is a median.
$\therefore QN = \frac{1}{2} QR$ ........... $(2)$
$\because BC = QR$ [Given]
$\therefore \frac{1}{2} BC = \frac{1}{2} QR$
$\Rightarrow BM = QN$ [From $(1)$ and $(2)$]
$(i)$ In $\Delta ABM$ and $\Delta PQN$,we have:
$AB = PQ$ [Given]
$AM = PN$ [Given]
$BM = QN$ [Proved above]
$\therefore \Delta ABM \cong \Delta PQN$ [$SSS$ congruence criterion]
$(ii)$ $\because \Delta ABM \cong \Delta PQN$
$\therefore$ Their corresponding parts are congruent $(CPCT)$.
$\Rightarrow \angle B = \angle Q$
Now,in $\Delta ABC$ and $\Delta PQR$,we have:
$AB = PQ$ [Given]
$\angle B = \angle Q$ [Proved above]
$BC = QR$ [Given]
$\therefore \Delta ABC \cong \Delta PQR$ [$SAS$ congruence criterion]

(N/A) Given: $BE \perp AC$ and $CF \perp AB$ such that $BE = CF$.
To prove: $\Delta ABC$ is an isosceles triangle,i.e.,$AB = AC$.
Proof:
In right-angled triangles $\Delta BEC$ and $\Delta CFB$:
$1$. $\angle BEC = \angle CFB = 90^\circ$ (Altitudes are perpendicular to the sides).
$2$. $BC = CB$ (Common hypotenuse).
$3$. $BE = CF$ (Given).
Therefore,by $RHS$ congruence rule,$\Delta BEC \cong \Delta CFB$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
$\angle BCE = \angle CBF$
This implies $\angle BCA = \angle CBA$.
In $\Delta ABC$,since the angles opposite to equal sides are equal,the sides opposite to equal angles are also equal.
Therefore,$AB = AC$.
Hence,$\Delta ABC$ is an isosceles triangle.

(N/A) Given: $ABC$ is an isosceles triangle with $AB = AC$. $AP \perp BC$.
To prove: $\angle B = \angle C$.
Proof:
In $\Delta ABP$ and $\Delta ACP$:
$1$. $\angle APB = \angle APC = 90^\circ$ (Since $AP \perp BC$)
$2$. $AB = AC$ (Given)
$3$. $AP = AP$ (Common side)
Therefore,by $RHS$ congruence criterion,$\Delta ABP \cong \Delta ACP$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$\angle B = \angle C$.

(A) Given that $\triangle ABC \cong \triangle PQR$. By the rule of correspondence in congruent triangles,the corresponding parts are equal:
$AB = PQ$,$BC = QR$,and $AC = PR$.
Also,the angles are equal: $\angle A = \angle P$,$\angle B = \angle Q$,and $\angle C = \angle R$.
We are given that $\triangle ABC$ is not congruent to $\triangle RPQ$. This implies that the correspondence $A \leftrightarrow R$,$B \leftrightarrow P$,and $C \leftrightarrow Q$ does not hold.
Let's check the options based on $\triangle ABC \cong \triangle PQR$:
Option $A$: $BC = PQ$. From the correspondence,$BC = QR$. Thus,$BC = PQ$ is generally not true.
Option $B$: $AC = PR$. This is true by correspondence.
Option $C$: $QR = BC$. This is true by correspondence.
Option $D$: $AB = PQ$. This is true by correspondence.
Therefore,the statement that is not true is $BC = PQ$.

(B) The criteria for congruence of triangles are $SAS$ (Side-Angle-Side),$ASA$ (Angle-Side-Angle),$SSS$ (Side-Side-Side),$AAS$ (Angle-Angle-Side),and $RHS$ (Right angle-Hypotenuse-Side).
$SSA$ (Side-Side-Angle) is not a valid criterion for the congruence of triangles because it does not uniquely determine a triangle.

(C) Given the side correspondences: $AB = QR$,$BC = PR$,and $CA = PQ$.
To determine the congruence,we map the vertices based on the equal sides:
$1$. Since $AB = QR$,vertex $A$ corresponds to $Q$ and vertex $B$ corresponds to $R$.
$2$. Since $BC = PR$,vertex $B$ corresponds to $P$ and vertex $C$ corresponds to $R$.
$3$. Since $CA = PQ$,vertex $C$ corresponds to $P$ and vertex $A$ corresponds to $Q$.
Matching the vertices: $A \leftrightarrow Q$,$B \leftrightarrow R$,and $C \leftrightarrow P$ is not correct based on the side $BC=PR$. Let's re-evaluate:
$A$ corresponds to $Q$,$B$ corresponds to $R$,and $C$ corresponds to $P$.
Looking at the sides: $AB$ corresponds to $QR$,$BC$ corresponds to $RP$,and $CA$ corresponds to $PQ$.
Thus,$\triangle ABC \cong \triangle QRP$.
Re-checking the options provided:
If $AB=QR, BC=PR, CA=PQ$,then the correspondence is $A \leftrightarrow Q, B \leftrightarrow R, C \leftrightarrow P$.
Therefore,$\triangle CBA \cong \triangle PRQ$ is correct because $CB=PR, BA=RQ, AC=QP$.

(D) According to the $SAS$ (Side-Angle-Side) congruence axiom,two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle.
Given: $AB = FD$ and $\angle A = \angle D$.
For the triangles to be congruent by $SAS$,the sides forming the angle $\angle A$ (which are $AB$ and $AC$) must be equal to the sides forming the angle $\angle D$ (which are $FD$ and $DE$).
Since $AB = FD$ is already given,we must have $AC = DE$ for the $SAS$ condition to be satisfied.

(A) To prove that $\triangle ABC \cong \triangle DEF$ (or a variation thereof) using the $SAS$ (Side-Angle-Side) congruence criterion,the angle must be included between the two given sides.
For $\triangle ABC$,the sides are $AB$ and $AC$,so the included angle is $\angle A$.
For $\triangle DEF$,the sides are $DE$ and $EF$,so the included angle is $\angle E$.
Therefore,if $\angle A = \angle E$,then by the $SAS$ congruence rule,$\triangle ABC \cong \triangle DEF$.

(NO) The two triangles are not necessarily congruent. For two triangles to be congruent by the $ASA$ (Angle-Side-Angle) criterion,the side must be included between the two angles. In $\triangle ABC$,the side included between $\angle A$ and $\angle B$ is $AB$. In $\triangle DEF$,the side included between $\angle D$ and $\angle E$ is $DE$. Since the given condition is $AB = EF$ instead of $AB = DE$,the triangles do not satisfy the $ASA$ congruence criterion. Therefore,they are not necessarily congruent.

(N/A) In triangles $ABC$ and $PQR$,we have:
$\angle A = \angle Q$ (Given)
$\angle B = \angle R$ (Given)
For the two triangles to be congruent by the $ASA$ (Angle-Side-Angle) congruence rule,the side included between the two angles must be equal.
In $\triangle ABC$,the side included between $\angle A$ and $\angle B$ is $AB$.
In $\triangle PQR$,the side included between $\angle Q$ and $\angle R$ is $QR$.
Therefore,for the triangles to be congruent,side $AB$ must be equal to side $QR$ $(AB = QR)$.

(N/A) In triangles $ABC$ and $PQR$,we are given:
$\angle A = \angle Q$ and $\angle B = \angle R$.
For the two triangles to be congruent by the $AAS$ (Angle-Angle-Side) congruence rule,the side corresponding to $BC$ in $\triangle ABC$ must be equal to the side corresponding to $RP$ in $\triangle PQR$.
Therefore,$BC = RP$.
This ensures that the two triangles are congruent by the $AAS$ congruence criterion.

(B) The statement is false. For two triangles to be congruent by the $SAS$ (Side-Angle-Side) congruence rule,the angle must be the included angle between the two sides. If the angle is not the included angle,the triangles are not necessarily congruent.

(A) The statement is true,provided that the side is the included side between the two angles ($ASA$ Congruence Rule) or the side is corresponding to the same relative position in both triangles ($AAS$ Congruence Rule).
If two angles and any side of one triangle are equal to the corresponding two angles and the corresponding side of another triangle,the triangles are congruent by the $AAS$ (Angle-Angle-Side) congruence criterion.
Therefore,the condition of correspondence is essential for the triangles to be congruent.

(N/A) In $\triangle PQS$ and $\triangle PRT$:
$1$. $PQ = PR$ (Given)
$2$. $\angle Q = \angle R$ (Given)
$3$. $\angle QPS = \angle RPT$ (Common angle $\angle P$)
Therefore,by the $ASA$ (Angle-Side-Angle) congruence criterion,$\triangle PQS \cong \triangle PRT$.

(N/A) Given: $BC \parallel AD$ and $BC = DA$.
Consider $\triangle BOC$ and $\triangle AOD$:
$1$. $\angle CBO = \angle DAO$ (Alternate interior angles,since $BC \parallel AD$)
$2$. $\angle BCO = \angle ADO$ (Alternate interior angles,since $BC \parallel AD$)
$3$. $BC = DA$ (Given)
By the $ASA$ (Angle-Side-Angle) congruence criterion,$\triangle BOC \cong \triangle AOD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
$OB = OA$ and $OC = OD$.
Therefore,$O$ is the midpoint of both the line segments $AB$ and $CD$.

(N/A) Given: $\triangle ABC$ with $AB = AC$.
$D$ is the mid-point of $AC$ and $E$ is the mid-point of $AB$.
To prove: $BD = CE$.
Proof: In $\triangle ABD$ and $\triangle ACE$:
$1$. $AB = AC$ (Given)
$2$. $\angle BAD = \angle CAE$ (Common angle)
$3$. $AD = AE$ (Since $AB = AC$,their halves are equal,i.e.,$\frac{1}{2}AB = \frac{1}{2}AC$)
By $SAS$ congruence criterion,$\triangle ABD \cong \triangle ACE$.
Therefore,$BD = CE$ by $CPCT$ (Corresponding Parts of Congruent Triangles).

(N/A) Given: $\triangle ABC$ in which $BD = CE$ and $AD = AE$.
To Prove: $\triangle ABD \cong \triangle ACE$.
Proof: In $\triangle ADE$,we have
$AD = AE$ [Given]
$\Rightarrow \angle ADE = \angle AED$ [Angles opposite to equal sides of a triangle are equal]
Now,$\angle ADB + \angle ADE = 180^{\circ}$ [Linear pair axiom]
$\angle AEC + \angle AED = 180^{\circ}$ [Linear pair axiom]
From these equations,we get:
$\angle ADB + \angle ADE = \angle AEC + \angle AED$
Since $\angle ADE = \angle AED$,we have $\angle ADB = \angle AEC$.
Now,in $\triangle ABD$ and $\triangle ACE$:
$AD = AE$ [Given]
$\angle ADB = \angle AEC$ [Proved above]
$BD = CE$ [Given]
So,by $SAS$ criterion of congruence,we have
$\triangle ABD \cong \triangle ACE$.
Hence,proved.

(N/A) Given: An equilateral triangle $CDE$ is formed on side $CD$ of square $ABCD$.
To prove: $\triangle ADE \cong \triangle BCE$.
Proof: In square $ABCD$,we have:
$AD = BC$ (Sides of a square are equal) ... $(1)$
$\angle ADC = \angle BCD = 90^{\circ}$ (Angles of a square) ... $(2)$
In equilateral triangle $CDE$,we have:
$DE = CE$ (Sides of an equilateral triangle are equal) ... $(3)$
$\angle CDE = \angle DCE = 60^{\circ}$ (Angles of an equilateral triangle) ... $(4)$
Adding $(2)$ and $(4)$:
$\angle ADC + \angle CDE = \angle BCD + \angle DCE$
$\angle ADE = \angle BCE$ ... $(5)$
Now,in $\triangle ADE$ and $\triangle BCE$:
$AD = BC$ (From $1$)
$\angle ADE = \angle BCE$ (From $5$)
$DE = CE$ (From $3$)
Therefore,by $SAS$ congruence criterion,$\triangle ADE \cong \triangle BCE$.

(N/A) Given: $BA \perp AC$,$DE \perp DF$,$BA = DE$,and $BF = EC$.
Step $1$: Establish the equality of the hypotenuses.
We are given $BF = EC$.
Adding $FC$ to both sides,we get:
$BF + FC = EC + FC$
$BC = EF$
Step $2$: Compare the two triangles.
In $\triangle ABC$ and $\triangle DEF$:
$1$. $\angle A = \angle D = 90^{\circ}$ (Given as $BA \perp AC$ and $DE \perp DF$)
$2$. $BA = DE$ (Given)
$3$. $BC = EF$ (Hypotenuse,proved above)
Step $3$: Conclusion.
By the $RHS$ (Right angle-Hypotenuse-Side) congruence rule,$\triangle ABC \cong \triangle DEF$.

(N/A) In $\triangle AMC$ and $\triangle BMD$,we have:
$\angle MAC = \angle MBD$ (Alternate interior angles,since $l \parallel m$)
$\angle AMC = \angle BMD$ (Vertically opposite angles)
$AM = BM$ (Given,as $M$ is the mid-point of $AB$)
Therefore,$\triangle AMC \cong \triangle BMD$ (By $ASA$ congruence rule)
Thus,$CM = DM$ (By $CPCT$)
Hence,$M$ is also the mid-point of $CD.$

(N/A) We have a right-angled $\Delta ABC$ such that $BCED$,$ACFG$,and $ABMN$ are squares on its sides $BC$,$CA$,and $AB$ respectively. Line segment $AX \perp DE$ is drawn such that it meets $BC$ at $Y$.
To prove that $\Delta MBC \cong \Delta ABD$:
In $\Delta ABD$ and $\Delta MBC$,we have:
$AB = MB$ (Sides of the square $ABMN$)
$BD = BC$ (Sides of the square $BCED$)
$\angle MBA = 90^\circ$ and $\angle CBD = 90^\circ$ (Angles of squares)
Therefore,$\angle MBA = \angle CBD = 90^\circ$.
Adding $\angle ABC$ to both sides:
$\angle MBA + \angle ABC = \angle CBD + \angle ABC$
$\angle MBC = \angle ABD$
Thus,by the $SAS$ (Side-Angle-Side) congruence criterion:
$\Delta MBC \cong \Delta ABD$.

(N/A) To prove: $\Delta FCB \cong \Delta ACE$
In $\Delta FCB$ and $\Delta ACE$:
$1$. $FC = AC$ (Sides of the same square $ACFG$)
$2$. $CB = CE$ (Sides of the same square $BCED$)
$3$. $\angle FCA = \angle BCE = 90^\circ$ (Angles of squares)
Adding $\angle ACB$ to both sides:
$\angle FCA + \angle ACB = \angle BCE + \angle ACB$
$\angle FCB = \angle ACE$
Therefore,by the $SAS$ (Side-Angle-Side) congruence criterion:
$\Delta FCB \cong \Delta ACE$

(N/A) This is the $ASA$ (Angle-Side-Angle) congruence criterion.
Let there be two triangles $\triangle ABC$ and $\triangle DEF$ such that $\angle B = \angle E$,$\angle C = \angle F$,and $BC = EF$.
We need to prove that $\triangle ABC \cong \triangle DEF$.
Case $1$: If $AB = DE$,then $\triangle ABC \cong \triangle DEF$ by $SAS$ congruence rule.
Case $2$: If $AB < DE$,take a point $P$ on $DE$ such that $DP = AB$. Join $PF$. In $\triangle ABC$ and $\triangle DPF$,$AB = DP$,$\angle B = \angle E$,and $BC = EF$. Thus,$\triangle ABC \cong \triangle DPF$ by $SAS$. This implies $\angle ACB = \angle DPF$. But we are given $\angle ACB = \angle DFE$. Therefore,$\angle DPF = \angle DFE$,which is only possible if $P$ coincides with $E$. Thus,$AB = DE$ and $\triangle ABC \cong \triangle DEF$.
Case $3$: If $AB > DE$,a similar argument shows $AB = DE$ and the triangles are congruent.

(N/A) The defect in the student's argument is that they used the $AAS$ (Angle-Angle-Side) congruence criterion,but the side $AD$ is not included between the angles $\angle B$ and $\angle ADB$ in $\triangle ABD$.
Correct proof:
In $\triangle ABD$ and $\triangle ACD$:
$AB = AC$ (Given)
$AD = AD$ (Common side)
$\angle ADB = \angle ADC = 90^{\circ}$ (Given $AD \perp BC$)
Therefore,by $RHS$ (Right angle-Hypotenuse-Side) congruence criterion:
$\triangle ABD \cong \triangle ACD$
Thus,$\angle BAD = \angle CAD$ $(CPCT)$.

(N/A) In $\triangle ABD$ and $\triangle CBD$,we have:
$AB = CB$ [Given]
$AD = CD$ [Given]
$BD = BD$ [Common side]
Therefore,$\triangle ABD \cong \triangle CBD$ [By $SSS$ congruence rule]
$\Rightarrow \angle ABD = \angle CBD$ [$CPCT$]
And $\angle ADB = \angle CDB$ [$CPCT$]
Hence,$BD$ bisects both the angles $ABC$ and $ADC$.

(N/A) In $\triangle ADC$ and $\triangle BEC$ we have:
$AC = BC$ [Given] ... $(1)$
$\angle ADC = \angle BEC = 90^{\circ}$ [Given as altitudes]
$\angle ACD = \angle BCE$ [Common angle]
Therefore,$\triangle ADC \cong \triangle BEC$ [By $AAS$ congruence rule]
Therefore,$CD = CE$ ... $(2)$ [$CPCT$]
Subtracting $(2)$ from $(1)$,we get:
$AC - CE = BC - CD$
$AE = BD$
Hence,proved.

(N/A) Given: Lines $l$ and $m$ intersect at point $O$. $P$ is a point on line $n$ passing through $O$ such that $PQ \perp l$ and $PR \perp m$,where $PQ = PR$.
To prove: $n$ is the bisector of $\angle QOR$.
Proof: In $\triangle OQP$ and $\triangle ORP$,we have:
$\angle OQP = \angle ORP = 90^{\circ}$ (Given that $P$ is equidistant from $l$ and $m$)
$OP = OP$ (Common side)
$PQ = PR$ (Given)
By $RHS$ (Right angle-Hypotenuse-Side) criterion of congruence,we have:
$\triangle OQP \cong \triangle ORP$
Therefore,$\angle QOP = \angle ROP$ (by $CPCT$ - Corresponding Parts of Congruent Triangles).
Thus,$n$ is the bisector of $\angle QOR$. Hence proved.

(N/A) Given: $A$ quadrilateral $ABCD$ such that $\angle 1 = \angle 2$ and $\angle 3 = \angle 4.$
To prove: $AB = AD$ and $CB = CD.$
Proof: In $\triangle ABC$ and $\triangle ADC$,we have:
$\angle 1 = \angle 2$ [Given]
$AC = AC$ [Common side]
$\angle 3 = \angle 4$ [Given]
So,by $ASA$ criterion of congruence,we have:
$\triangle ABC \cong \triangle ADC$
Therefore,$AB = AD$ [by $CPCT$]
And $CB = CD$ [by $CPCT$]
Hence,proved.

(N/A) Given: $A$ quadrilateral $ABCD$ in which $AB = AD$ and $CB = CD$.
To prove: $AC$ is the perpendicular bisector of $BD$.
Proof: In $\triangle ABC$ and $\triangle ADC$,we have:
$AB = AD$ (Given)
$BC = CD$ (Given)
$AC = AC$ (Common side)
So,by $SSS$ congruence criterion,we have:
$\triangle ABC \cong \triangle ADC$
Therefore,$\angle 1 = \angle 2$ (by $CPCT$)
Now,in $\triangle AOB$ and $\triangle AOD$,we have:
$AB = AD$ (Given)
$\angle 1 = \angle 2$ (Proved above)
$AO = AO$ (Common side)
So,by $SAS$ congruence criterion,we have:
$\triangle AOB \cong \triangle AOD$
Therefore,$BO = DO$ (by $CPCT$)
And $\angle 3 = \angle 4$ (by $CPCT$)
But,$\angle 3 + \angle 4 = 180^{\circ}$ (Linear pair axiom)
Since $\angle 3 = \angle 4$,we have:
$\angle 3 + \angle 3 = 180^{\circ}$
$2\angle 3 = 180^{\circ}$
$\angle 3 = 90^{\circ}$
Since $BO = DO$ and $\angle 3 = 90^{\circ}$,$AC$ is the perpendicular bisector of $BD$.
Hence,proved.

(N/A) Given: $\angle PXR = \angle SXQ$
Subtracting $\angle QXR$ from both sides:
$\angle PXR - \angle QXR = \angle SXQ - \angle QXR$
$\therefore \angle PXQ = \angle SXR \quad \dots(1)$
Now,in $\Delta XPQ$ and $\Delta XSR$:
$XP = XS$ (Given)
$XQ = XR$ (Given)
$\angle PXQ = \angle SXR$ [From $(1)$]
$\therefore \Delta XPQ \cong \Delta XSR$ (By $SAS$ congruence criterion)
$\therefore PQ = SR$ (By $CPCT$)

(N/A) In rectangle $ABCD$,
$AB = DC$ (Opposite sides of a rectangle are equal)
and $\angle B = \angle C = 90^{\circ}$ (Each angle of a rectangle is $90^{\circ}$).
Also,$E$ is the midpoint of $BC$,so $BE = CE$.
Now,consider $\Delta ABE$ and $\Delta DCE$:
$AB = DC$ (Proved above)
$\angle B = \angle C = 90^{\circ}$ (Proved above)
$BE = CE$ (Given that $E$ is the midpoint of $BC$)
By the $SAS$ (Side-Angle-Side) congruence criterion,$\Delta ABE \cong \Delta DCE$.
Since the triangles are congruent,their corresponding parts are equal.
Therefore,$AE = DE$ (by $CPCT$ - Corresponding Parts of Congruent Triangles).

(N/A) Given:
$1$. $PN \perp PQ$ and $QM \perp PQ$.
$2$. $X$ is the midpoint of $PQ$,so $PX = QX$.
$3$. $X$ is the midpoint of $MN$,so $NX = MX$.
To prove: $\triangle PNX \cong \triangle QMX$.
Proof:
In $\triangle PNX$ and $\triangle QMX$:
$1$. $PX = QX$ (Given,$X$ is the midpoint of $PQ$).
$2$. $NX = MX$ (Given,$X$ is the midpoint of $MN$).
$3$. $\angle PNX = \angle QMX$ (Since $PN \parallel QM$ because both are perpendicular to $PQ$,and $MN$ is a transversal,these are alternate interior angles).
Alternatively,using $SAS$ congruence criterion:
$1$. $PX = QX$ (Given).
$2$. $\angle P = \angle Q = 90^\circ$ (Given).
$3$. $NX = MX$ (Given).
Therefore,by $RHS$ congruence criterion (or $SAS$),$\triangle PNX \cong \triangle QMX$.

(N/A) Consider $\triangle APM$ and $\triangle BPN$.
$1$. $\angle MAP = \angle NBP = 90^{\circ}$ (Given that $AM \perp AB$ and $BN \perp AB$).
$2$. $AP = BP$ ($P$ is the midpoint of $AB$,given).
$3$. $\angle APM = \angle BPN$ (Vertically opposite angles).
Therefore,by the $ASA$ (Angle-Side-Angle) congruence criterion,$\triangle APM \cong \triangle BPN$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
- $AM = BN$ (Hence proved).
- $PM = PN$ (Since $PM$ and $PN$ are corresponding sides).
Since $PM = PN$,$P$ is the midpoint of $MN$ (Hence proved).

(N/A) Given: $PQ = ST$,$QU = TR$,$\angle PQT = 90^\circ$ and $\angle STQ = 90^\circ$.
To prove: $PR = SU$.
Proof:
Consider $\triangle PQR$ and $\triangle STU$.
$1$. $PQ = ST$ (Given)
$2$. $\angle PQR = \angle STU = 90^\circ$ (Given)
$3$. $QR = QU + UR$ and $TU = TR + UR$.
Since $QU = TR$,we have $QR = TR + UR = TU$.
Thus,$QR = TU$.
By $SAS$ congruence criterion,$\triangle PQR \cong \triangle STU$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$PR = SU$.

(N/A) Let $PQ$ be a line segment and $L$ be its perpendicular bisector. Let $M$ be a point on $L$ such that $M$ does not lie on $PQ$. Let $L$ intersect $PQ$ at point $O$. Since $L$ is the perpendicular bisector of $PQ$,we have $PO = OQ$ and $\angle MOP = \angle MOQ = 90^{\circ}$. Consider $\triangle MOP$ and $\triangle MOQ$. In these triangles: $1$. $PO = OQ$ (Given). $2$. $\angle MOP = \angle MOQ = 90^{\circ}$ (Given). $3$. $MO = MO$ (Common side). By the $SAS$ (Side-Angle-Side) congruence criterion,$\triangle MOP \cong \triangle MOQ$. Since the triangles are congruent,their corresponding parts are equal $(CPCT)$. Therefore,$MP = MQ$. Thus,point $M$ is equidistant from $P$ and $Q$.

(N/A) In $\triangle ABC$,$AB = AC$ (Given).
$\therefore \angle ABC = \angle ACB$ ($\because$ Angles opposite to equal sides are equal).
Since $P$ lies on $BC$ and $Q$ lies on $BC$,we have $\angle ABP = \angle ACQ$.
Now,in $\triangle ABP$ and $\triangle ACQ$:
$AB = AC$ (Given)
$BP = CQ$ (Given)
$\angle ABP = \angle ACQ$ (Proved above)
So,by $SAS$ congruence criterion,$\triangle ABP \cong \triangle ACQ$.
$\therefore AP = AQ$ (by $CPCT$).
Now,in $\triangle APQ$,since $AP = AQ$,$\Delta APQ$ is an isosceles triangle.

(N/A) Given: $AB$ and $CD$ bisect each other at $P$. This implies $PA = PB$ and $PC = PD$.
However,the problem states $PA = PD$ and $PB = PC$.
Consider $\triangle APC$ and $\triangle BPD$:
$1$. $PA = PD$ (Given)
$2$. $\angle APC = \angle BPD$ (Vertically opposite angles)
$3$. $PC = PB$ (Given)
By $SAS$ (Side-Angle-Side) congruence criterion,$\triangle APC \cong \triangle BPD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AC = BD$.

(N/A) In $\Delta SPQ$ and $\Delta RQP$:
$PS = QR$ (Given)
$QS = PR$ (Given)
$PQ = QP$ (Common side)
Therefore,by $SSS$ congruence criterion,$\Delta SPQ \cong \Delta RQP$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$(1) \angle PSQ = \angle QRP$ and $(2) \angle SPQ = \angle RQP$.

(N/A) Given: $AB \perp BQ$,$PQ \perp QB$,$AC = PR$ and $BR = QC$.
Since $AB \perp BQ$,$\angle ABQ = 90^{\circ}$. Thus,$\angle ABC = 90^{\circ}$.
Since $PQ \perp QB$,$\angle PQB = 90^{\circ}$. Thus,$\angle PQR = 90^{\circ}$.
Given $BR = QC$.
Adding $RC$ to both sides,we get $BR + RC = QC + RC$.
Therefore,$BC = QR$......$(1)$
In $\Delta ABC$ and $\Delta PQR$:
$AC = PR$ (Given)
$\angle ABC = \angle PQR = 90^{\circ}$ (Proved above)
$BC = QR$ [From $(1)$]
By $RHS$ congruence criterion,$\Delta ABC \cong \Delta PQR$.
Therefore,$\angle BAC = \angle QPR$ (by $CPCT$).