Solution

SSA is not a criterion for congruence of triangles.

Hence, the correct option is (C).

Solution

Given:

$AB=QR, BC=PR$ and $CA=PQ$, then

The vertices are one-one corresponding that is $P$ corresponding to $C, Q$ to $A$ and $R$ to $B$, which is written as:

$P \leftrightarrow C, Q \leftrightarrow A, R \leftrightarrow B$

Under that correspondence, we have:

$\triangle C B Q \cong \triangle P R Q$

Hence, the correct option is (B).

Solution

According to the question, triangle $ABC$ is:

$AB=AC$ [Given]

So, $\angle C=\angle B$ [Angles opposite to equal sides are equal]

Given: $\angle B=50^{\circ}$. So, $\angle C=50^{\circ}$

Hence, the correct option is (B).

Solution

In triangle $ABC$ :

$BC=AB$ [given]

$\angle A=\angle C$ [Since, angles opposite to equal sides are equal]

$\angle B=80^{\circ}$

Therefore, $\angle A+\angle B+\angle C=180^{\circ}$

$\angle A+80^{\circ}-\angle A=180^{\circ}$

$2 \angle A=100^{\circ}$

$\angle A=\dfrac{100^{\circ}}{2}$

$\angle A=50^{\circ}$

Hence, the correct option is (C).

Solution

In triangle $PQR$ :

$\angle R=\angle P$ [Given]

$PQ=QR$ [Sides opposite to equal angles are equal]

Now, $QR=4 ~cm$, therefore, $PQ=4 ~cm$.

Therefore, the length of the $PQ$ is $4 ~cm$

Hence, the correct option is (A).

Solution

In triangle $ADC$,

Ext. $\angle A D B>$ Int. opp $\angle D A C$

$\angle A D B>\angle B A D$

[Because: $\angle B A D=\angle D A C]$ $A B>B D$ [Side opposite to greater angle is longer]

Hence, the correct option is (B).

Solution

Given: $\triangle ABC \cong \triangle FDE$ and $AB=5 ~cm, \angle B=40^{\circ}$ and $\angle A=80^{\circ}$

$ \begin{matrix} DF=AB & {[By \mathrm{CPCT]}} \\ DF=5 ~cm & \\ \angle E=\angle C & {[By \mathrm{CPCT]}} \\ \angle E=\angle C=180^{\circ}-(\angle A+\angle B) & \text{ [By angle sum property of a triangle ABC] } \\ \angle E=180^{\circ}-(80^{\circ}+40^{\circ}) & \\ \angle E=60^{\circ} & \end{matrix} $

Solution

Sum of any two sides of a triangle is greater than third side. So, third side of the triangle cannot be $3.4 ~cm$ because then,

$1.5~ ~cm+3.4 ~cm=4.9 ~cm<$ third side $[5 ~cm]$

Hence, the correct option is (D).

Solution

Given: In triangle $PQR$,

$\angle R>\angle Q$

$PQ>PR$ [side opposite to greater angle is longer]

Hence, the correct option is (B).

Solution

In triangle $ABC$,

$AB=AC \quad$ [Given]

$\angle C=\angle B \quad$ [Angles opposite to equal sides are equal]

So, in triangle $ABC$ is an isosceles triangle.

$\angle B=\angle Q \quad$ [Given]

$\angle C=\angle P$

$\angle P=\angle Q \quad$ [Since, $\angle C=\angle B]$

$QR=PR$ [Sides opposite to equal angles are equal]

So, in triangle $PQR$ is also an isosceles triangle.

Hence, both triangle are isosceles but not congruent.

Hence, the correct option is (A).

Solution

Given, in $\triangle ABC$ and $\triangle DEF, AB=DF$ and $\angle A=\angle D$

As we know that, two triangles will be congruent by ASA rule, if two angles and the included side of one triangle are equal to the two angles and the included side of other triangle.

Since, $AC=DE$

Hence, the correct option is (B).

Solution

Given: in $\triangle ABC$ and $\triangle PQR, \angle A=\angle Q$ and $\angle B=\angle R$

Now, the triangle $ABC$ and $PQR$ will be congruent if $AB=QR$ by $ASA$ congruence rule.

Solution

Given: In triangle $ABC$ and $PQR$,

$\angle A=\angle Q$ and $\angle B=\angle R \quad$ [given]

$BC=RP[$ For the triangle to be congruent]

Hence, it will be congruent by AAS congruence rule.

Solution

Angle must be the included angles. Hence, this statement is not true.

Solution

As we know that, the sum of any two sides of the triangle is always greater than the third side.

Solution

As we know that, the sum of any two sides of the triangle is always greater than the third side. So,

$4 ~cm$ and $3 ~cm=4 ~cm+3 ~cm=7 ~cm$ that is equal to the length of third side that is $7 ~cm$.

Therefore, this is not possible to construct a triangle with length of sides $4 ~cm, 3 ~cm$ and $7 ~cm$.

Solution

It is false that $BC=QR$ because $BC=PQ$ as $\triangle A B C \cong \triangle R P Q$.

Solution

It is true, $PR=EF$ because this is the corresponding sides of triangle $PQR$ and triangle $EDF$.

Solution

In triangle $PQR$,

$ \begin{aligned} \angle Q & =180^{\circ}-(\angle P+\angle R) \\ & =180^{\circ}-(70^{\circ}+30^{\circ}) \\ & =180^{\circ}-100^{\circ} \\ & =80^{\circ} \end{aligned} $

Now, in triangle $PQR$, angle $Q$ is larger and side opposite to greater angle is longer.

Therefore, PR is the longer side.

Solution

In triangle $ABD$,

$AB+BD>AD$

$AC+CD>AD \ldots$ (II) [Sum of the lengths of any two sides of a triangle must be greater that the third side]

Adding (I) and (II), get:

$AB+BD+CD+AC>2 AD$

$AB+BC+CA>2 AD \quad[BD=CD$ as $AD$ is median of triangle $ABC]$

Solution

To prove: $AB+BC+AC>2 AM$

Proof: We know that sum of any two side of a triangle is greater than the third side,

Now, in triangle $A B M$,

$AB+BM>Am \ldots(I)$

And, in triangle ACM,

$AC+~CM>AM \ldots$…(II)

Adding (I) and (II), get:

$AB+BM+AC+~CM>2 AM$

$AB+(BM+~CM)+AC>2 AM$

$AB+BC+AC>2 AM$

Hence, it is true that the perimeter of the triangle is greater than $2 AM$.

Solution

We know that sum of any two side of a triangle is greater than the third side. So,

$9 ~cm+7 ~cm=16 ~cm<17 ~cm$

Hence, it is not possible to construct a triangle.

Solution

Yes, that is possible to construct a triangle with lengths of its sides as $8 ~cm, 7 ~cm$ and $4 ~cm$ because the sum of any two side of a triangle is greater than the third side.

Solution

Given:

$ABC$ is an isosceles triangle with $AB=AC$ and $BD$ and $CE$ are its two medians.

To prove: $BD=CE$

Proof: in triangle $ABC$,

$AB=AC \quad$ [Given]

$\dfrac{1}{2} A B=\dfrac{1}{2} A C$

$AE=AD[D$ is the mid-point of $AC$ and $E$ is the mid-point of $AB]$

Now, in triangle $ABD$ and triangle $ACE$,

$AB=AC \quad[$ Given]

$\angle A=\angle A \quad[$ Common angle]

$AE=AD$ [above proved]

Now, by SAS criterion of congruence, get:

$\triangle A B D \cong \triangle A C E$

$BD=CE \quad[CPCT]$

Hence, proved.

Solution

Given in $\triangle A B C~$ in which $BD=CE$ and $AD=AE$

To prove that $\triangle ABD \cong \triangle ACE$

Proof: $AD=AE$ $\angle A D E=\angle A E D$ $\angle A D B+\angle A D E=180^{\circ}$ [Given] [Since, angle opposite to equal sides are equal] … (I) $\angle A D B=180^{\circ}-\angle A D E$ [Linear pair axiom] $\angle A D B=180^{\circ}-\angle A E D$ [From equation (i)]

In triangle $ABD$ and triangle $ACE$, $\angle A D B=\angle A E C$ [Since, $\angle A E C+\angle A E D=180^{\circ}$, linear pair axiom] $BD=CE$ [Given] $AD=AE$ [Given] $\triangle A B D \cong \triangle A C E$ [By SAS congruence rule]

Solution

Given in figure triangle $CDE$ is an equilateral triangle formed on a side $CD$ of a square $ABCD$. To proof that $\triangle A D E \cong \triangle B C E$

Proof: In triangle $ADE$ and triangle $BCE$,

$DE=CE \quad$ [Sides of an equilateral triangle]

$\angle A D E=\angle B C E$

$\angle A D C=\angle B C D=90^{\circ}$ and $\angle E D C=\angle E C D=60^{\circ}$

$ \begin{matrix} \angle A D E=90^{\circ}+60^{\circ}=150^{\circ} \text{ and } \angle B C E=90^{\circ}+60^{\circ}=150^{\circ} \\ AD=BC & \text{ [Sides of a square] } \\ \triangle A D E \cong \triangle B C E & \text{ [By SAS congruence rule] } \end{matrix} $

Solution

See in the figure,

$BA \perp AC, DE \perp DF$ such that $BA=DE$ and $BF=EC$

To proof that $\triangle ADC \cong \triangle DEF$

Proof:

$BF=EC \quad[$ Given]

Now, adding $CF$ both sides, get:

$B F+C F=E C+C F$

$BC=EF$

In triangle $ABC$ and triangle $DEF$, $\angle A=\angle D=90^{\circ}$ $[BA \perp AC$ and $DE \perp DF]$ $BC=EF$ [from eq. (I)] $BA=DE$ [Given] $\triangle A B C \cong \triangle D E F$ [By RHS congruence rule]

Solution

In triangle $PSR, Q$ is a point on the side $SR$ such that $PQ=PR$.

To proof that PS $>$ PQ

Proof: In triangle PRQ,

$PQ=PR$

[Given]

$\angle R=\angle P Q R \quad$…(I) [Angle opposite to equal sides are equal] $\angle P Q R>\angle S$

… (II) [Exterior angle of a triangle is greater than each of the opposite

interior angle]

Now, from equation (I) and (II), get:

$\angle R>\angle S$

$PS>PR$ [side opposite to greater angle is longer]

$PS>PQ \quad[PQ=PR]$

Solution

Given in triangle $PQR, S$ is any point on side $QR$.

To proof that $PQ+QR+RP>2 PS$

Proof: In triangle PQS,

$PQ+QS>PS$ (i) [Sum of two side of a triangle is greater than the third side]

Now, similarly in triangle PRS,

$SR+RP>PS$ (ii) [Sum of two side of a triangle is greater than the third side]

Adding equation (I) and (II), get:

$PQ+QS+SR+RP>2 PS$

$PQ+(QS+SR)+RP>2 PS$

$PQ+QR+RP>2 PS \quad[QR=QS+SR]$

Solution

Given in triangle $A B C, D$ is any point on side $A C$ such that $A B=A C$.

To proof that $CD<BD$ or $BD>CD$

To proof:

$AC=AB$

[Given]

In triangle $ABC$ and triangle $DBC$,

$\angle A B C>\angle D B C \quad[\angle D B C$ is a internal angle of $\angle B]$

$\angle A C B>\angle D B C \quad$ [From equation (I)]

$BD>CD \quad$ [Side opposite to greater angle is longer]

$CD<BD$

Solution

See in the figure, $l || m$ and $M$ is the mid-point of a line segment $AB$.

To proof that $MC=MD$

Proof: $l || m$ [Given]

$\angle B A C=\angle A B D$ [Alternate interior angles]

$\angle A M C=\angle B M D$ [Vertical opposite angle]

In triangle $AMC$ and triangle $BMD$,

$\angle B A C=\angle A B D$

$AM=BM$

$\angle A M C=\angle B M D$

$MC=MD$ [Proved above]

[Given]

[By ASA congruence rule]

[ By CPCT]

Solution

Given in the question, bisectors of the angles $B$ and $C$ of an isosceles triangle $ABC$ with $AB=$ $AC$ intersect each other at $O$. Now $BO$ is produced to a point $M$.

In triangle $ABC$,

$AB=AC$

$\angle A B C=\angle A C B \quad$ [Angle opposite to equal sides of a triangle are equal]

$\dfrac{1}{2} \angle A B C=\dfrac{1}{2} \angle A C B$

That is $\angle 1=\angle 2$ [Since, $BO$ and $Co$ are bisectors of $\angle B$ and $\angle C$ ]

In triangle $OBC$, Ext. $\angle M O C=\angle 1+\angle 2$ [Exterior angle of a triangle is equal to the sum of interior opposite angles]

Ext. $\angle M O C=2 \angle 1 \quad[\angle 1=\angle 2]$

Hence, $\angle M O C=\angle A B C$.

Solution

In triangle $ABC$,

$AB=AC$

So, $\angle B=\angle C$ [Angle opposite to equal sides of a triangle are equal]

$\dfrac{1}{2} \angle B=\dfrac{1}{2} \angle C$

In triangle $OBC$,

$\angle 1=\dfrac{1}{2} \angle B$

And, $\angle 2=\dfrac{1}{2} \angle 2$

$\angle D B C+\angle 1+\angle O B A=180^{\circ} \quad$ [ABD is a straight line]

In triangle $OBC$,

$\angle 1+\angle 2+\angle B O C=180^{\circ}$

$2 \angle 1+\angle B O C=180^{\circ}$

$[\angle 1=\angle 2] \ldots$ (II)

From equation (I) and (II), get:

$\angle D B A+2 \angle 1=2 \angle 1+\angle B O C$ $\angle D B C=\angle B O C$

Solution

In triangle $ACD$,

Ext. $\angle A D B>\angle D A C$ [Exterior angle of a triangle is greater than either of the interior opposite angle]

$\angle A D B>\angle B A D$

Since, in triangle $ABD$,

$\angle A D B>\angle B A D$

Hence, $AB>BD$. [In a triangle, side opposite to greater angle is longer]

Solution

In triangle $ABC$,

$ \begin{aligned} & AB=AC \\ & \angle C=\angle B \quad \ldots \text{ (I) [Angles opposite to equal sides of a triangle are equal] } \\ & BC=AC \\ & \angle A=\angle B \quad \ldots \text{ (II) } \\ & \angle A+\angle B+\angle C=180^{\circ} \text{ [Angle sum property of a triangle] } \\ & \angle A+\angle A+\angle A=180^{\circ} \text{ [From equation (I) and (II)] } \\ & \angle A=\dfrac{180^{\circ}}{3} \\ & \angle A=60^{\circ} \end{aligned} $

Solution

Let $AB$ intersect $LM$ at $O$.

To prove: $AO=BO$.

Proof: $\angle i=\angle r$

$\angle B=\angle r$ [Corresponding angle]

…(I) [Angle of incidence $=$ Angle of reflection $]$

Now,

Since, from equation (I), (II) and (III), get:

$\angle B=\angle A$

$\angle B C O=\angle A C O$

In triangle $BOC$ and triangle $AOC$, get:

$ \begin{matrix} \angle 1=\angle 2 & {[\text{ Each }=90^{\circ}]} \\ OC=OC & {[\text{ Common side }]} \\ \angle B C O=\angle A C O & \text{ [Prove above] } \\ \triangle B O C \cong \triangle A O C & \text{ [ASA congruence rule] } \\ \text{ Hence, } AO=BQ & {[CPCT]} \end{matrix} $

Solution

In $\triangle ADB$ and $\triangle ADC$, get:

$\angle A D B=\angle A D C$ $[$ Each equal to $90^{\circ}]$

$AB=AC$ $[$ Given]

$AD=AD$ $[$ Common side $]$

Now, by RHS criterion of congruence, get:

$\triangle A D B \cong \triangle A D C$

So, $\angle B A D=\angle C A D \quad[CPCT]$

Hence, proved.

Solution

Given in the question, $P$ is a point on the bisector of $\angle ABC$. If the line through $P$, parallel to $BA$ meet $BC$ at $Q$.

To prove: BPQ is an isosceles triangle.

Proof: $\angle 1=\angle 2$ …(I) [BP is the bisector of $\angle A B C$ ]

$PQ$ is parallel to BA and BP cuts them. So,

$\angle 1=\angle 3$

[Alternate interior angles as $PQ || AB$ ]

$\angle 2=\angle 3$

[Proved above] $PQ=BQ$

Hence, BPQ is an isosceles triangle.

[Sides opposite to equal angle are equal]

Solution

In triangle $ABC$ and triangle $CBD$,

$ \begin{matrix} AB=BC & {[\text{ Given }]} \\ AD=CD & {[\text{ Given }]} \\ BD=BD & {[\text{ Common side }]} \end{matrix} $

So, $\triangle A B C \cong \triangle C B D \quad$ [By SSS congruence rule] $\angle 1=\angle 2 \quad[CPCT]$

And, $\angle 3=\angle 4$

Hence, $BD$ bisects both the angle $ABC$ and $ADC$.

Solution

Given in the question, $ABC$ is a right triangle with $AB=AC$. Bisector of $\angle A$ meets $BC$ at $D$.

To prove that $BC=2 AD$

Proof: In right triangle $ABC$,

$AB=AC$

[Given]

$BC$ is hypotenuse. So, $\angle B A C=90^{\circ}$

Now, in triangle $CAD$ and triangle $BAD$, get: $AC=AB$ [Given] $\angle 1=\angle 2$ [AD is the bisector of $\angle A$ ] $AD=AD~$[Common side]

Now, by SAS criterion of congruence, get:

$\triangle C A D \cong \triangle B A D$

$CD=BD$

[CPCT]

$AD=BD=CD \ldots$. (I)

vertices of a triangle]

Now, $BC=BD+CD$

$B C=A D+A D$

[Using (I)]

$BC=2 AD$

[Mid-point of hypotenuse of a rt. Triangle is equidistant from the three

Hence, proved.

Solution

Given in the question, A square of $ABCD$ and $OA=OB=AB$.

To prove that triangle OCD is an isosceles triangle.

Proof: In triangle $ABCD$, $\angle 1=\angle 2$ … (I) [Each equal to $.90^{\circ}]$

In triangle $OAB$,

$\angle 3=\angle 4 \quad \ldots$ (II) [Each equal to $60^{\circ}$ ]

Now, subtracting equation (II) from equation (I), get: $\angle 1-\angle 3=\angle 2-\angle 4$

$\angle 5=\angle 6$

In triangle $DAO$ and triangle $CBO$,

$AD=BC \quad$ [Given]

$\angle 5=\angle 6 \quad$ [Proved above]

$OA=OB \quad$ [Given]

So, by SAS criterion of congruence, get:

$\triangle D A O \cong \triangle C B O$

$OD=OC$

Now, in triangle OCD is an isosceles triangle.

Hence, proved.

Solution

Given in the question, $ABC$ and $DBC$ are two triangles on the same base $BC$ such that $A$ and $D$ lie on the opposite sides of $BC, AB=AC$ and $DB=DC$.

To proof that $AD$ is the perpendicular bisector of $BC$ that is $OB=OC$.

Proof: In triangle $BAD$ and triangle $CAD$,

$AB=AC \quad[$ Given]

$BD=CD \quad[$ Given]

$AD=AD \quad[$ Common side $]$

Now, by SSS criterion of congruence,

$\triangle B A D \cong \triangle C A D$

So, $\angle 1=\angle 2 \quad[CPCT]$

Now, in triangle BAO and triangle CAO,

$AB=AC$ [Given]

$\angle 1=\angle 2 \quad$ [Proved above] $AO=AO \quad[$ Common side $]$

So, by SAS criterion of congruence,

$\triangle B A O \cong \triangle C A O$

Since, $BO=CO$

And, $\angle 3=\angle 4[CPCT]$

$\angle 3+\angle 4=180^{\circ} \quad$ [Linear pair axiom]

$\angle 3+\angle 3=180^{\circ}$

$[CPCT]$

$ \begin{aligned} 2\angle 3 & =180^{\circ} \\ \angle 3 & =\dfrac{180^{\circ}}{2} \\ \angle 3 & =90^{\circ} \end{aligned} $

Therefore, $AD$ is perpendicular to bisector of $BC$.

Hence, proved.

Solution

In triangle $ADC$ and triangle $BEC$,

$AC=BC$

[Given]…(I)

$\angle A D C=\angle B E C$

[Each is $90^{\circ}$ ]

$\angle A C D=\angle B C E$

[Common angle]

So, $\triangle A D C \cong \angle B E C$

[By SSS congruence rule]

$CE=CD$

…(II) $[CPCT]$

Now, Subtracting equation (II) from (I), get:

$AC-AE=BC-CD$

$AE=BD$

Hence, proved.

Solution

Given in triangle $ABC$ with median $AD$,

To proof:

$AB+AC>2 AD$

$AB+BC>2 AD$

$BC+AC>2 AD$

Producing $AD$ to $E$ such that $DE=AD$ and join $EC$.

Proof: In triangle $ADB$ and triangle EDC,

$AD=ED \quad$ [By construction]

$\angle 1=\angle 2 \quad$ [Vertically opposite angles are equal]

$DB=DC \quad[$ Given]

So, by SAS criterion of congruence,

$\triangle A D B \cong \triangle E D C$

$AB=EC \quad[CPCT]$

And, $\angle 3=\angle 4[CPCT]$

Again, in triangle AEC,

$AC+CE>AE$ [Sum of the lengths of any two sides of a triangle must be greater than the third side]

$AC+CE>AD+DE$

$AC+CE>AD+AD[AD=DE]$

$AC+CE>2 AD$

$AC+AB>2 AD \quad[$ Because $AB=CE]$

Hence proved.

Similarly, $AB+BC>2 AD$ and $BC+AC>2 AD$.

Solution

Given in the question, A quadrilateral,

To prove that $AB+BC+CD<2(BD+AC)$

Proof: In triangle AOB,

$OA+OB>AB$

than the third side]

…(I) [Sum of the lengths of any two sides of a triangle must be greater

In triangle $BOC$,

$OB+OC>BC \ldots$ (II) [Same reason]

In triangle COD,

$OC+OD>CD \ldots$ (III) $\quad[$ Same reason]

In triangle DOA,

$OD+OA>DA \ldots$ (IV) [Same reason]

Now, adding equation (I), (II), (III) and (IV), get:

$OA+OB+OB+OC+OD+OD+OA>AB+BC+CD+DA$

$2(OA+OB+OC+OD)>AB+BC+CD+DA$

$2{(OA+OC)+(OB+OD)}>AB+BC+CD+DA$

$2(AC+BD)>AB+BC+CD+DA$

$AB+BC+CD+DA<2(BD+AC)$

Hence, proved.

Solution

Given in the question, a quadrilateral $ABCD$.

To proof that $AB+BC+CD+DA>AC+BD$.

Proof: In triangle $ABC$,

$AB+BC>AC \ldots$…(I) [Sum of the lengths of any two sides of a triangle must be greater than the third side]

In triangle $BCD$,

$BC+CD>BD$

…(II) [Sum of the lengths of any two sides of a triangle must be greater

than the third side]

In triangle $CDA$,

$AD+DA>AC$…(III) [Sum of the lengths of any two sides of a triangle must be greater than the third side]

Similarly, in triangle $DAB$,

$AD+AB>BD$

than the third side]

…(IV) [Sum of the lengths of any two sides of a triangle must be greater

Now, adding equation (I), (II), (III) and (IV), get:

$AB+BC+BC+CD+CD+DA+AD+AB>AC+BD+AC+BD$

$2 AB+2 BC+2 CD>2 AC+2 BD$

$2(AB+BC+CD+DA)>2(AC+BD)$

$AB+BC+CD+DA>AC+BD$

Hence, proved.

Solution

Given: $D$ is the mid-point of side $AC$.

To proof: $\angle A B C=90^{\circ}$

Proof: $AD=DC$

And, $BD=\dfrac{1}{2} A C=A D \quad[D$ is the mid-point of side $AC]$

$BD=AD=DC$

In triangle $ABD$,

$BD=AD$

$\angle 1=\angle 2 \quad$… (I) [Angles opposite to equal sides are equal]