3 Marks Question

Question 13 Marks

Two sides $AB$ and $BC$ and median $AM$ of the $\triangle ABC$ are respectively equal to side $PQ$ and $QR$ and median $PN $of $\text{PQR} ($See figure$)$. Show that:
$i. \triangle ABM \cong \triangle PQN$
$ii. \triangle ABC \cong \triangle PQR$

Answer

$AM$ is the median of $\triangle A B C$.
$\therefore B M=M C=\frac{1}{2} B C \ldots \text { (i) }$
$P N$ is the median of $\triangle P Q R$.
$\therefore QN=NR=\frac{1}{2} QR \text {. }$
Now $BC = QR [$ Given $]$
$\Rightarrow \frac{1}{2} B C=\frac{1}{2} Q R$
$\therefore BM=QN \ldots \text { (iii) }$
$i$. Now in $\triangle A B M$ and $\triangle P Q N$,
$AB = PQ [$ Given $]$
$AM = PN [$ Given]
$BM = QN [$ From eq.$(iii)]$
$\therefore \triangle ABM \cong \triangle PQN [$By $\text{SSS}$ congruency$]$
$\Rightarrow \angle B=\angle Q [$By $\text {C.P.C.T.}] ...(iv)$
$ii.$ In $\triangle A B C$ and $\triangle P Q R$,
$AB=PQ[$Given $]$
$\angle B=\angle Q [$Prove above$]$
$\therefore P R=Q R [$Given$]$
$A B C \cong P Q R [$By $\text{SAS}$ congruency$]$

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Question 23 Marks

$AD$ is an altitude of an isosceles $\triangle ABC$ in which $AB = AC$. Show that
$i. AD$ bisects $BC$
$ii. AD$ bisects $\angle A.$

Answer

Given : $A D$ is an altitude of an isosceles $\triangle A B C$ in which $A B=A C$ To prove :
$i. A D$ bisects $B C$
$ii. A D$ bisects $\angle A$
Proof:
$i.$ In right $\triangle A D B$ and right $\triangle A D C$,
$A B=A C \ldots [$Given$]$
Side $A D=$ Side $A D \ldots [$Common$]$
$\therefore \triangle ADB \cong \triangle ADC \ldots [\text{RHS}$ Rule$]$
$\therefore B D=C D \ldots [\text{c.p.c.t.}]$
$\therefore AD$ bisects $BC .$
$ii. \triangle ADB \cong \triangle ADC \ldots [$As proved above$]$
$\therefore \angle B A D=\angle C D A \ldots [\text{c.p.c.t.}]$
$\therefore AD$ bisects $\angle A$

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Question 33 Marks

Show that the angles of an equilateral triangle are $60^\circ $ each.

Answer

Let $ABC$ is an equilateral triangle. We know that all the sides of an equilateral triangle are equal.
$\therefore AB = BC = CA ....(1)$

To prove :- $\angle A = \angle B = \angle C = 60^\circ $
Proof :-
In $\Delta ABC$ we have:-
AB = AC [from (1)]
$\Rightarrow \angle C = \angle B ....(2)$
[$\because$ Angles opposite to equal sides of a triangle are equal]
Again from $(1),$
$BC = AC$
$\Rightarrow \angle A = \angle B .....(3)$
[$\because$ Angles opposite to equal sides of a triangle are equal] .
From $(2) \& (3) ;$
$\Rightarrow \angle A = \angle B=\angle C ....(4)$
Now,
$\angle A + \angle B + \angle C = 180 ^ { \circ }$ [$\because$ Angle sum property of a triangle]
$\Rightarrow \angle A + \angle A + \angle A = 180 ^ { \circ }$ [From$ (4) ]$
$\Rightarrow 3 \angle A = 180 ^ { \circ }$|
$\Rightarrow \angle A = \frac { 180 ^ { \circ } } { 3 } = 60 ^ { \circ }$
$\therefore \angle A = \angle B = \angle C = 60 ^ { \circ }$[ from $(4) ]. $
Hence, each angle of an equilateral triangle is equal to $60^\circ .$

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Question 43 Marks

$\Delta ABC$ is an isosceles triangle in which $AB = AC$. Side $BA$ is produced to $D$ such that $AD = AB$ (See figure). Show that $\angle BCD$ is a right angle.

Answer

From figure & according to the question, in $\Delta ABC,$
$AB = AC ..........(1)$
$\Rightarrow \angle ACB = \angle ABC ...(2)$ [Angles opposite to equal sides are equal]
Again given,$ AD = AB$
But $AB = AC$ [ from $(1) ]$
$\therefore AD = AB = AC$
$\Rightarrow AD = AC..... (3)$
Now in $\Delta ADC,$
$AD = AC$ [ from $(3) ]$
$\Rightarrow \angle A D C = \angle A C D …(4)$ [Angles opposite to equal sides are equal]
In $\Delta$ BCD,
$\angle A B C + \angle B C D + \angle C D A = 180 ^ { \circ }$[ Angle sum property ]
$\Rightarrow \angle A C B + \angle B C D + \angle C D A = 180 ^ { \circ }$[ Because $\angle ACB = \angle ABC$, from $(2) ]$
$\Rightarrow \angle A C B + \angle A C B + \angle A C D + \angle C D A = 180 ^ { \circ }$ [Because $\angle B C D = \angle A C B + \angle A C D$ ]
$\Rightarrow 2 \angle A C B + \angle A C D + \angle C D A = 180 ^ { \circ }$
$\Rightarrow 2 \angle A C B + \angle A C D + \angle A C D = 180 ^ { \circ }$ [ Because $\angle ADC = \angle ACD$, see $(4) ]]$
$\Rightarrow 2 \angle A C B + 2 \angle A C D = 180 ^ { \circ }$
$\Rightarrow 2 ( \angle A C B + \angle A C D ) = 180 ^ { \circ }$ [ Taking out $2$ common ]
$\Rightarrow 2 \angle B C D = 180 ^ { \circ }$ [ Because, $\angle A C D + \angle A C B = \angle B C D$ ]
$\Rightarrow$ $\angle B C D = 90 ^ { \circ }$
Hence $\angle BCD$ is a right angle.

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Question 53 Marks

In an isosceles triangle $ABC$ , with $AB = AC$, the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$ . Join A to $O$ . Show that $O B=O C$ and $A O$ bisects $\angle A$.

Answer

Given: In $\triangle A B C, A B=A C$, the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Construction: Joint $A$ to $O$

To prove: $O B=O C$ and $A O$ bisects $A$.
Proof: $A B=A C \ldots$ [Given]
$\therefore \angle B=\angle C \ldots[\angle$ s opposite to equal side of a $\triangle]$
$\therefore \frac{1}{2} \angle B=\frac{1}{2} \angle C$
$\therefore \angle OBC =\angle OCB \ldots$ [As BO bisects $\angle B$ and $CO$ bisects $\angle C ]$
$\therefore O B=O C \ldots$ [Sides opposite to equal $\angle$ s of a $\triangle$ ]
In $\triangle O A B$ and $\triangle O A C$,
$A B=A C \ldots$ [Given]
$O B=O C \ldots$ [As proved above]
$OA = OA \ldots$ [Common]
$\therefore \triangle OAB \cong \triangle OAC . . .$ [By $SSS$ property]
$\therefore \angle OAB = \angle OAC . . . $[c.p.c.t.]
$\therefore AO$ bisects $\angle A$

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Question 63 Marks

$AB$ is a line segment and $P$ is its mid$-$point. $D $and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE$ and $\angle EPA = \angle DPB.$
Show that:
$i. \triangle DAP \cong \triangle EBP$
$ii. AD = BE$

Answer

Given: $AB$ is a line segment and $P$ is its mid$-$point.
$D$ and $E$ are points on the same side of $AB$
such that $ \angle BAD = \angle ABE$ and $\angle EPA = \angle DPB.$
To prove:
$i. \text{DDAP} \cong \text{DEBP}$
$ii. AD = BE$
Proof :$(ii)$
$\angle EPA = \angle DPB ...[$Given$]$
$\angle EPA + \angle EPD = \angle EPD + \angle DPB ...[$Adding $\angle EPD$ to both sides$]$
$∠APD = ∠BPE ...(1)$
In $DDAP$ and $DEBP$
$\angle DAP = \angle EBP ...[$Given$]$
$AP = BP ...[$As $P$ is the mid$-p$oint of the line $AB]$
$\angle APD = \angle BPE ...[$From $(1)]$
$\therefore DDAP \cong DEBP$ proved$ ...[\text{ASA}$ property$] ...(2)$
$(i)$ As $DDAP \cong DEBP ...[$From $(2)]$
$\therefore AD = BE ... [\text{c.p.c.t.}]$

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Question 73 Marks

Angles opposite to equal sides of an isosceles triangle are equal.This result can be proved in many ways. One of the proofs is given here.

Answer

Proof: We are given an isosceles triangle $A B C$ in which $A B=A C$. We need to prove that $\angle \mathrm{B}=\angle \mathrm{C}$.
Let us draw the bisector of $\angle \mathrm{A}$ and let $\mathrm{D}$ be the point of intersection of this bisector of $\angle \mathrm{A}$ and $\mathrm{BC}$ (see Fig. 7.25).
In $\triangle \mathrm{BAD}$ and $\triangle \mathrm{CAD}$,
$
\begin{aligned}
\mathrm{AB} =\mathrm{AC} \quad (Given) \\
\angle \mathrm{BAD} =\angle \mathrm{CAD} \quad \text{(By construction)} \\
\mathrm{AD} =\mathrm{AD} \quad (Common) \\
So, \quad \triangle \mathrm{BAD} \cong \triangle \mathrm{CAD} \text{(By SAS rule)}
\end{aligned}
$
So, $\angle \mathrm{ABD}=\angle \mathrm{ACD}$, since they are corresponding angles of congruent triangles.
So, $\angle \mathrm{B}=\angle \mathrm{C}$

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