3 Marks Question

Question 13 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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Question 23 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 PQR (See figure). Show that:

$\triangle ABM \cong \triangle PQN$
$\triangle ABC \cong \triangle PQR$

Answer

AM is the median of $\triangle$ABC.
$\therefore $ BM = MC = $\frac{1}{2}$ BC ...(i)
PN is the median of $\triangle$PQR.
$\therefore $ QN = NR = $\frac{1}{2}$ QR ...(ii)
Now BC = QR [Given] $\Rightarrow$ $\frac{1}{2} BC =\frac{1}{2} QR$
$\therefore $ BM = QN ...(iii)

Now in $\triangle$ABM and $\triangle$PQN,
AB = PQ[Given]
AM = PN[Given]
BM = QN[From eq.(iii)]
$\therefore $ $\triangle$ABM $\cong$ $\triangle$PQN [By SSS congruency]
$\Rightarrow$ $\angle$B = $\angle$Q [By C.P.C.T.] ...(iv)
In $\triangle$ABC and $\triangle$PQR,
AB = PQ [Given]
$\angle$B = $\angle$Q [Prove above]
$\therefore $ PR = QR [Given]
ABC $\cong$ PQR [By SAS congruency]

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Question 33 Marks

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

AD bisects BC
AD bisects $\angle$A.

Answer

Given : AD is an altitude of an isosceles triangle ABC in which AB = AC
To prove :

AD bisects BC
AD bisects $\angle$A

Proof :

In right $\triangle$ADB and right $\triangle$ADC,
AB = AC . . . . [Given]
Side AD = Side AD . . . . [Common]
$\therefore$ $\triangle$ADB $\cong$ $\triangle$ADC . . . [RHS Rule]
$\therefore$ BD = CD . . . [c.p.c.t.]
$\therefore$ AD bisects BC.
$\triangle$ADB $\cong$ $\triangle$ADC . . . [As proved above]
$\therefore$ $\angle$BAD = $\angle$CDA . . . [c.p.c.t.]
$\therefore$ AD bisects $\angle$A

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Question 43 Marks

Show that the angles of an equilateral triangle are 60° 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 :- $\angle A = \angle B = \angle C = 60°$
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°.

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Question 53 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 63 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 OB = OC and AO bisects $\angle$A.

Answer

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

To prove: OB = OC and AO bisects A.
Proof : AB = AC . . . . [Given]
$\therefore$ $\angle$B = $\angle$C . . . [$\angle$s opposite to equal side of a $\triangle$]
$\therefore$ $\frac{1}{2}$$\angle$B = $\frac{1}{2}$$\angle$C
$\therefore$ ∠OBC = ∠OCB . . . [As BO bisects $\angle$B and CO bisects $\angle$C]
$\therefore$ OB = OC . . . [Sides opposite to equal $\angle$s of a $\triangle$ ]
In $\triangle$OAB and $\triangle$OAC,
AB = AC . . . .[Given]
OB = OC . . . .[As proved above]
OA = OA . . . .[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 73 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:

$\triangle$DAP $\cong$ $\triangle$EBP
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:

DDAP $\cong$ DEBP
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-point of the line AB]
$\angle$APD = $\angle$BPE ...[From (1)]
$\therefore$ DDAP $\cong$ DEBP proved ...[ASA property] ...(2)
(i) As DDAP $\cong$ DEBP ...[From (2)]
$\therefore$ AD = BE ...[c.p.c.t.]

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