Mix Examples - Triangles Questions in English

(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) In $\triangle ABC$,we are given that $AB = AC$.
Since the angles opposite to equal sides of a triangle are equal,we have $\angle C = \angle B$.
Given that $\angle B = 50^{\circ}$.
Therefore,$\angle C = 50^{\circ}$.

(A) In $\triangle ABC$,we are given that $BC = AB$.
Since the sides opposite to equal angles are equal,the angles opposite to equal sides are also equal.
Therefore,$\angle A = \angle C$.
We know that the sum of all interior angles of a triangle is $180^{\circ}$.
So,$\angle A + \angle B + \angle C = 180^{\circ}$.
Substituting the given values: $\angle A + 80^{\circ} + \angle A = 180^{\circ}$.
$2 \angle A + 80^{\circ} = 180^{\circ}$.
$2 \angle A = 180^{\circ} - 80^{\circ}$.
$2 \angle A = 100^{\circ}$.
$\angle A = 100^{\circ} / 2 = 50^{\circ}$.

(B) In $\triangle PQR$,we are given that $\angle R = \angle P$.
According to the property of triangles,the sides opposite to equal angles are equal.
Therefore,the side opposite to $\angle R$ (which is $PQ$) must be equal to the side opposite to $\angle P$ (which is $QR$).
Thus,$PQ = QR$.
Given that $QR = 4 \, cm$,it follows that $PQ = 4 \, cm$.

(C) In $\triangle ADC$,the exterior angle $\angle ADB$ is greater than the interior opposite angle $\angle DAC$.
Since $AD$ is the bisector of $\angle BAC$,we have $\angle BAD = \angle DAC$.
Substituting this into the inequality,we get $\angle ADB > \angle BAD$.
In $\triangle ABD$,since the side opposite to the greater angle is longer,we have $AB > BD$.

(D) Given that $\triangle ABC \cong \triangle FDE$.
$1$. In $\triangle ABC$,the sum of angles is $180^{\circ}$. Therefore,$\angle C = 180^{\circ} - (80^{\circ} + 40^{\circ}) = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
$2$. Since $\triangle ABC \cong \triangle FDE$,the corresponding parts of congruent triangles are equal $(CPCT)$.
$3$. The correspondence is $A \leftrightarrow F$,$B \leftrightarrow D$,and $C \leftrightarrow E$.
$4$. Therefore,$AB$ corresponds to $FD$ (or $DF$). Thus,$DF = AB = 5 \, cm$.
$5$. Also,$\angle C$ corresponds to $\angle E$. Thus,$\angle E = \angle C = 60^{\circ}$.
$6$. Comparing this with the given options,$DF = 5 \, cm$ and $\angle E = 60^{\circ}$ is correct.

(A) According to the triangle inequality theorem,the sum of any two sides of a triangle must be greater than the third side,and the difference between any two sides must be less than the third side.
Let the third side be $x$.
Then,$5 - 1.5 < x < 5 + 1.5$,which simplifies to $3.5 < x < 6.5$.
Checking the options:
$A) 3.4$ is not in the range $(3.5, 6.5)$ because $3.4 < 3.5$.
$B) 3.8$ is in the range.
$C) 4.1$ is in the range.
$D) 3.6$ is in the range.
Therefore,the length of the third side cannot be $3.4\, cm$.

(B) In $\triangle PQR,$ we are given that $\angle R > \angle Q.$
According to the theorem,in any triangle,the side opposite to the greater angle is longer.
The side opposite to $\angle R$ is $PQ$ and the side opposite to $\angle Q$ is $PR.$
Since $\angle R > \angle Q,$ it follows that $PQ > PR.$

(C) Given: In $\triangle ABC$,$AB = AC$. Therefore,$\angle B = \angle C$ (Angles opposite to equal sides are equal).
Given: $\angle C = \angle P$ and $\angle B = \angle Q$.
Since $\angle B = \angle C$,we have $\angle Q = \angle P$.
In $\triangle PQR$,since $\angle Q = \angle P$,the sides opposite to these angles must be equal,so $PR = QR$. Thus,$\triangle PQR$ is an isosceles triangle.
However,we do not have enough information (such as a side length or a specific angle measure) to prove that $\triangle ABC \cong \triangle PQR$. For example,the triangles could have different sizes while maintaining the same angle relationships.
Therefore,both triangles are isosceles,but they are not necessarily congruent.

(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) According to the triangle inequality theorem,the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side.
In this case,the lengths of the sides are $4 \, cm, 3 \, cm$,and $7 \, cm$.
Calculating the sum of the two smaller sides: $4 \, cm + 3 \, cm = 7 \, cm$.
Since the sum of the two sides $(7 \, cm)$ is equal to the third side $(7 \, cm)$ and not greater than it,the triangle inequality condition is not satisfied.
Therefore,it is not possible to construct a triangle with these side lengths.

(B) It is false to say that $BC = QR$.
Since $\triangle ABC \cong \triangle RPQ$,the corresponding parts of congruent triangles are equal $(CPCT)$.
By matching the vertices,we have $A \leftrightarrow R$,$B \leftrightarrow P$,and $C \leftrightarrow Q$.
Therefore,the side $BC$ corresponds to the side $PQ$.
Thus,$BC = PQ$,not $QR$.

(A) Yes,it is true that $PR = EF$.
Since $\Delta PQR \cong \Delta EDF$,the corresponding parts of congruent triangles are equal $(CPCT)$.
The vertices correspond as follows: $P \leftrightarrow E$,$Q \leftrightarrow D$,and $R \leftrightarrow F$.
Therefore,the side $PR$ corresponds to the side $EF$,which implies $PR = EF$.

(PR) In $\triangle PQR$,the sum of all interior angles is $180^{\circ}$.
$\angle Q = 180^{\circ} - (\angle P + \angle R)$
$\angle Q = 180^{\circ} - (70^{\circ} + 30^{\circ}) = 180^{\circ} - 100^{\circ} = 80^{\circ}$.
Comparing the angles,we have $\angle Q > \angle P > \angle R$ $(80^{\circ} > 70^{\circ} > 30^{\circ})$.
According to the property of triangles,the side opposite to the largest angle is the longest side.
The angle $\angle Q$ is the largest angle,and the side opposite to $\angle Q$ is $PR$.
Therefore,$PR$ is the longest side of $\triangle PQR$.

(N/A) In $\triangle ABD$,we have:
$AB + BD > AD$ $\ldots(1)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
In $\triangle ADC$,we have:
$AC + CD > AD$ $\ldots(2)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
Adding $(1)$ and $(2)$,we get:
$AB + BD + CD + AC > 2AD$
Since $AD$ is the median of $\triangle ABC$,$BD = CD$. Therefore,$BD + CD = BC$.
Substituting this into the inequality,we get:
$AB + BC + CA > 2AD$
Yes,the statement is true.

(A) We have to prove that $AB + BC + AC > 2\, AM$.
According to the triangle inequality theorem,the sum of any two sides of a triangle is greater than the third side.
In $\triangle ABM$,we have:
$AB + BM > AM$ $\ldots (1)$
In $\triangle ACM$,we have:
$AC + CM > AM$ $\ldots (2)$
Adding equations $(1)$ and $(2)$,we get:
$AB + BM + AC + CM > AM + AM$
$AB + AC + (BM + CM) > 2\, AM$
Since $M$ is a point on $BC$,$BM + CM = BC$. Substituting this into the inequality:
$AB + AC + BC > 2\, AM$
Thus,the perimeter of the triangle $ABC$ is indeed greater than $2\, AM$.

(N/A) No,it is not possible to construct a triangle with side lengths $9 \, cm$,$7 \, cm$,and $17 \, cm$.
According to the triangle inequality theorem,the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
Here,the sum of the two smaller sides is:
$9 \, cm + 7 \, cm = 16 \, cm$
Since $16 \, cm < 17 \, cm$,the sum of the two sides is not greater than the third side.
Therefore,a triangle cannot be formed with these side lengths.

(A) Yes,it is possible to construct a triangle with side lengths $8\,cm, 7\,cm$ and $4\,cm$.
According to the triangle inequality theorem,the sum of any two sides of a triangle must be greater than the third side.
Checking the conditions:
$8 + 7 = 15 > 4$
$8 + 4 = 12 > 7$
$7 + 4 = 11 > 8$
Since all three conditions are satisfied,the triangle can be constructed.

(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: In $\triangle PQR$,$PQ > PR$ and $QS$,$RS$ are bisectors of $\angle Q$ and $\angle R$ respectively.
Since $PQ > PR$,the angle opposite to the longer side is greater. Therefore,$\angle R > \angle Q$.
Since $QS$ and $RS$ are bisectors of $\angle Q$ and $\angle R$ respectively,we have:
$\angle SQR = \frac{1}{2} \angle Q$
$\angle SRQ = \frac{1}{2} \angle R$
Since $\angle R > \angle Q$,it follows that $\frac{1}{2} \angle R > \frac{1}{2} \angle Q$.
Therefore,$\angle SRQ > \angle SQR$.
In $\triangle SQR$,since $\angle SRQ > \angle SQR$,the side opposite to the greater angle is longer. Thus,$SQ > SR$.

(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) Given: $PQ = PR$.
To prove: $PS > PQ$.
Proof: In $\triangle PQR$,we have:
$PR = PQ$ (Given).
Therefore,$\angle PQR = \angle PRQ$ (Angles opposite to equal sides of a triangle are equal).
In $\triangle PQS$,$\angle PQR$ is an exterior angle.
We know that the exterior angle of a triangle is greater than each of the interior opposite angles.
Therefore,$\angle PQR > \angle S$.
Since $\angle PQR = \angle PRQ$,we have $\angle PRQ > \angle S$.
In $\triangle PSR$,since $\angle PRQ > \angle S$,the side opposite to $\angle PRQ$ must be greater than the side opposite to $\angle S$.
Therefore,$PS > PR$.
Since $PR = PQ$,we can substitute $PQ$ for $PR$.
Thus,$PS > PQ$.
Hence proved.

(N/A) Given: $A$ point $S$ on side $QR$ of $\triangle PQR$.
To prove: $PQ + QR + RP > 2 \, PS$.
Proof: In $\triangle PQS$,we have:
$PQ + QS > PS$ ..... $(1)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
Now,in $\triangle PSR$,we have:
$RS + RP > PS$ ..... $(2)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
Adding $(1)$ and $(2)$,we get:
$PQ + QS + RS + RP > 2 \, PS$
Since $QS + RS = QR$,we have:
$PQ + QR + RP > 2 \, PS$
Hence,proved.

(N/A) In $\triangle ABC$,we have $AB = AC$ (Given).
Therefore,$\angle ABC = \angle ACB$ (Angles opposite to equal sides of a triangle are equal).
Since $D$ is a point on $AC$,$\angle DBC < \angle ABC$.
Substituting $\angle ACB$ for $\angle ABC$,we get $\angle DBC < \angle ACB$.
Since $\angle ACB$ is the same as $\angle DCB$,we have $\angle DBC < \angle DCB$.
In $\triangle BCD$,since $\angle DCB > \angle DBC$,the side opposite to the greater angle is longer.
Therefore,$BD > CD$ or $CD < BD$.

(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) Given: In $\triangle ABC$,$AB = AC$. $BO$ and $CO$ are the angle bisectors of $\angle B$ and $\angle C$ respectively,intersecting at $O$. $BO$ is produced to $M$.
Proof:
In $\triangle ABC$,since $AB = AC$,we have $\angle ABC = \angle ACB$ (Angles opposite to equal sides are equal).
Since $BO$ and $CO$ are bisectors of $\angle B$ and $\angle C$,we have:
$\angle OBC = \frac{1}{2} \angle ABC$
$\angle OCB = \frac{1}{2} \angle ACB$
Since $\angle ABC = \angle ACB$,it follows that $\angle OBC = \angle OCB$.
In $\triangle OBC$,$\angle MOC$ is an exterior angle at vertex $O$ when side $BO$ is produced to $M$.
By the Exterior Angle Property of a triangle,the exterior angle is equal to the sum of the two interior opposite angles.
Therefore,$\angle MOC = \angle OBC + \angle OCB$.
Since $\angle OBC = \angle OCB$,we can write:
$\angle MOC = \angle OBC + \angle OBC = 2 \angle OBC$.
Substituting $\angle OBC = \frac{1}{2} \angle ABC$:
$\angle MOC = 2 \times (\frac{1}{2} \angle ABC) = \angle ABC$.
Hence,$\angle MOC = \angle ABC$.

(N/A) In $\triangle ABC$,we have $AB = AC$.
Therefore,$\angle ABC = \angle ACB$ [Since angles opposite to equal sides of a triangle are equal].
Since $BO$ and $CO$ are bisectors of $\angle B$ and $\angle C$ respectively,we have $\angle OBC = \frac{1}{2} \angle ABC$ and $\angle OCB = \frac{1}{2} \angle ACB$.
Since $\angle ABC = \angle ACB$,it follows that $\angle OBC = \angle OCB$.
In $\triangle OBC$,the sum of angles is $180^{\circ}$,so $\angle BOC + \angle OBC + \angle OCB = 180^{\circ}$.
Substituting $\angle OCB = \angle OBC$,we get $\angle BOC + 2 \angle OBC = 180^{\circ}$,which implies $\angle BOC = 180^{\circ} - 2 \angle OBC$.
Now,consider the external angle adjacent to $\angle ABC$. Let $D$ be a point on the extension of $AB$. The external angle is $\angle CBD = 180^{\circ} - \angle ABC$.
Since $\angle ABC = 2 \angle OBC$,we have $\angle CBD = 180^{\circ} - 2 \angle OBC$.
Comparing the two expressions,we conclude that $\angle BOC = \angle CBD$.

(N/A) Let $\angle BAD = \angle 1$ and $\angle CAD = \angle 2$. Since $AD$ is the bisector of $\angle BAC$,we have $\angle 1 = \angle 2$.
In $\triangle ACD$,$\angle ADC$ is an exterior angle at vertex $D$.
We know that an exterior angle of a triangle is equal to the sum of the two interior opposite angles.
Thus,$\angle ADC = \angle CAD + \angle ACD = \angle 2 + \angle ACD$.
Since $\angle ACD > 0$,it follows that $\angle ADC > \angle 2$.
Substituting $\angle 2 = \angle 1$,we get $\angle ADC > \angle 1$.
Now,consider $\triangle ABD$. In this triangle,the angle opposite to side $AB$ is $\angle ADB$ (which is $\angle ADC$) and the angle opposite to side $BD$ is $\angle BAD$ (which is $\angle 1$).
Since $\angle ADC > \angle 1$,the side opposite to the greater angle is longer.
Therefore,$AB > BD$.

(N/A) Produce $CB$ to a point $D$ such that $BC = BD$ and join $AD$.
In $\Delta ABC$ and $\Delta ABD$,we have:
$BC = BD$ (By construction)
$AB = AB$ (Common side)
$\angle ABC = \angle ABD = 90^{\circ}$ (Given and by construction)
Therefore,$\Delta ABC \cong \Delta ABD$ by $SAS$ congruence rule.
So,$\angle CAB = \angle DAB$ by $CPCT$. Let $\angle BAC = x$,then $\angle CAB = x$ and $\angle DAB = x$.
Thus,$\angle CAD = \angle CAB + \angle BAD = x + x = 2x$.
Since $\angle BCA = 2 \angle BAC$,and $\angle BAC = x$,then $\angle BCA = 2x$.
In $\Delta ACD$,$\angle CAD = 2x$ and $\angle ACD = 2x$.
Since two angles are equal,the sides opposite to them are equal,so $AC = CD$.
Also,since $\Delta ABC \cong \Delta ABD$,$AC = AD$ by $CPCT$.
Thus,$\angle ADC = \angle ACD = 2x$.
Since all three angles of $\Delta ACD$ are $60^{\circ}$ (as $2x + 2x + 2x = 180^{\circ} \implies x = 30^{\circ}$),$\Delta ACD$ is an equilateral triangle.
Therefore,$AC = CD = AD$.
Since $CD = BC + BD$ and $BC = BD$,we have $CD = BC + BC = 2BC$.
Hence,$AC = 2BC$.

(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) We are given a point $D$ on side $BC$ of a $\Delta ABC$ such that $\angle BAD = \angle CAD$ and $BD = CD$. We are to prove that $AB = AC$.
Produce $AD$ to a point $E$ such that $AD = DE$ and then join $CE$.
Now,in $\Delta ABD$ and $\Delta ECD$,we have:
$BD = CD$ (Given)
$AD = DE$ (By construction)
$\angle ADB = \angle EDC$ (Vertically opposite angles)
Therefore,$\Delta ABD \cong \Delta ECD$ ($SAS$ congruence rule).
So,$AB = EC$ and $\angle BAD = \angle CED$ $(CPCT)$ ... $(1)$
Also,$\angle BAD = \angle CAD$ (Given) ... $(2)$
From $(1)$ and $(2)$,we get $\angle CAD = \angle CED$.
In $\Delta ACE$,since $\angle CAD = \angle CED$,the sides opposite to these angles must be equal.
Therefore,$AC = EC$ ... $(3)$
From $(1)$ and $(3)$,we have $AB = AC$.
Hence,$\Delta ABC$ is an isosceles triangle.

(N/A) Produce $QS$ to intersect $PR$ at $T$.
From $\triangle PQT$,we have:
$PQ + PT > QT$ (The sum of any two sides of a triangle is greater than the third side)
$PQ + PT > SQ + ST$ ..... $(1)$
From $\triangle TSR$,we have:
$ST + TR > SR$ ..... $(2)$
Adding $(1)$ and $(2)$,we get:
$PQ + PT + ST + TR > SQ + ST + SR$
$PQ + PT + TR > SQ + SR$
Since $PT + TR = PR$,we have:
$PQ + PR > SQ + SR$
Therefore,$SQ + SR < PQ + PR$.

(A) In an equilateral triangle,all three sides are equal,i.e.,$AB = BC = AC$.
Since the angles opposite to equal sides of a triangle are equal,we have $\angle A = \angle B = \angle C$.
Let each angle be $x$.
By the angle sum property of a triangle,the sum of all interior angles is $180^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
$x + x + x = 180^{\circ}$.
$3x = 180^{\circ}$.
$x = \frac{180^{\circ}}{3} = 60^{\circ}$.
Thus,all the angles of an equilateral triangle are $60^{\circ}, 60^{\circ}, 60^{\circ}$.

(N/A) Let $AB$ intersect $LM$ at $O$. We have to prove that $AO = BO$.
Now,$\angle i = \angle r$ ... $(1)$
[$\because$ Angle of incidence = Angle of reflection]
$\angle B = \angle i$ [Corresponding angles] ... $(2)$
And $\angle A = \angle r$ [Alternate interior angles] ... $(3)$
From $(1)$,$(2)$ and $(3)$,we get $\angle B = \angle A$.
In $\triangle BOC$ and $\triangle AOC$,we have:
$\angle 1 = \angle 2 = 90^{\circ}$ [Given]
$OC = OC$ [Common side]
$\angle B = \angle A$ [Proved above]
Therefore,$\triangle BOC \cong \triangle AOC$ [$AAS$ congruence rule].
Hence,$AO = BO$ [$CPCT$].

(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) We have to prove that $\triangle BPQ$ is an isosceles triangle.
Given that $BP$ is the bisector of $\angle ABC$,we have:
$\angle 1 = \angle 2$ .....$(1)$
Since $PQ \parallel BA$ and $BP$ is a transversal,the alternate interior angles are equal:
$\angle 1 = \angle 3$ .....$(2)$
From equations $(1)$ and $(2)$,we get:
$\angle 2 = \angle 3$
In $\triangle BPQ$,since $\angle 2 = \angle 3$,the sides opposite to these equal angles must be equal:
$PQ = BQ$
Since two sides of $\triangle BPQ$ are equal,$\triangle BPQ$ is an isosceles triangle.

(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) Given: $A$ right-angled triangle $ABC$ where $AB = AC$ and the bisector of $\angle A$ meets $BC$ at $D$.
To prove: $BC = 2 AD$.
Proof: In $\triangle ABC$,since $AB = AC$ and it is a right-angled triangle,$\angle BAC = 90^{\circ}$.
In $\triangle CAD$ and $\triangle BAD$:
$AC = AB$ (Given)
$\angle CAD = \angle BAD$ (Since $AD$ is the bisector of $\angle A$)
$AD = AD$ (Common side)
By $SAS$ congruence criterion,$\triangle CAD \cong \triangle BAD$.
Therefore,$CD = BD$ (by $CPCT$).
Since $AD$ is the angle bisector of the right angle in an isosceles right triangle,$AD$ is also the median to the hypotenuse.
In a right-angled triangle,the median to the hypotenuse is half the length of the hypotenuse.
Thus,$AD = \frac{1}{2} BC$.
Multiplying both sides by $2$,we get $2 AD = BC$.
Hence,$BC = 2 AD$ is proved.