2 Marks Questions

Question 12 Marks

ABC is an isosceles triangle with AB = AC. Draw AP $\perp$ BC to show that $\angle$B = $\angle$C.

Answer

Given: ABC is an isosceles triangle with AB = AC.
To Prove : $\angle$B = $\angle$C
Construction: Draw AP $\perp$ BC
Proof: In right triangle APB and right triangle APC.
AB = AC . . . . [given]
AP = AP . . . .[Common]
$\therefore$ $\triangle$APB $\cong$ $\triangle$APC . . . [RHS rule]
$\therefore$ $\angle$ABP = $\angle$ACP . . .[c.p.c.t.]
$\therefore$ $\angle$B = $\angle$C

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Question 22 Marks

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that a triangle ABC is isosceles.

Answer

Given: BE and CF are two equal altitudes of a triangle ABC.
To Prove: $\triangle$ABC is isosceles.
Proof : In right $\triangle$BEC and right $\triangle$CFB
side BE = side CF ...[Given]
BC = CB ...[Common]
$\triangle$BEC = $\triangle$CFB ...[By RHS rule]
$\therefore$ ∠BCE = ∠CBF ...[c.p.c.t.]
$\therefore$ AB = AC ...[Sides opposite to equal angles of a triangle are equal]
$\therefore$ $\triangle$ABC is isosceles.

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Question 32 Marks

ABC is a right-angled triangle in which $\angle $A = ${90^ \circ }$ and AB = AC. Find $\angle $B and $\angle $C.

Answer

ABC is a right triangle in which,

AB = AC
$ \Rightarrow $ $\angle$C =$\angle$ B [ opposite angles of equal side are equal] .....(i)
We know that, in ABC, $\angle$A + $\angle $B + $\angle $C =$180^\circ $ [ Angle sum property of triangle]
$ \Rightarrow $ ${90^ \circ }$+$\angle $B + $\angle $B = $180^\circ $ [$\angle $ A = ${90^ \circ }$(given) and $\angle $B = C (from eq. (i)]
$ \Rightarrow $ 2$\angle $B = $180^\circ $- ${90^ \circ }$
$ \Rightarrow $ 2$\angle $B = ${90^ \circ }$
$ \Rightarrow$ $\angle$B = ${45^ \circ }$
Also $\angle $C = ${45^ \circ }$ [As $\angle $B = $\angle $C]

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Question 42 Marks

ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.

Answer

Given : ABC and DBC are two isosceles triangles on the same base BC.
To Prove : ∠ABD = ∠ACD.
Proof : As ABC is an isosceles triangle on the base BC
$\therefore$ ∠ABC = ∠ACB . . . .(1)
As ABC is an isosceles triangle on the base BC
$\therefore$ ∠DBC = ∠DCB . . . (2)
Adding the corresponding sides of (1) and (2)
∠ABC + ∠DBC = ∠ACB + ∠DCB
$\Rightarrow$ ∠ABD = ∠ACD

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Question 52 Marks

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that $\triangle \mathrm { ABE } \cong \triangle \mathrm { ACF },$ AB = AC i.e. $\triangle$ABC is an isosceles triangle.

Answer

Given : ABC is a triangle in which altitude BE and CF to side AC and AB are equal.
To Prove : $\triangle \mathrm { ABE } \cong \triangle \mathrm { ACF }$

AB = AC i.e. $\triangle$ABC is an isosceles triangle.
Proof : BE = CF ...... [Given]
∠BAE = ∠CAF ...... [Common]
∠AFB = ∠AFC ....... [Each 90^o]
$\therefore$ $\triangle \mathrm { ABE } \cong \triangle \mathrm { ACF }$ ........ [By AAS property]
$\triangle A B E \cong \triangle A C F$ ....... [As proved]
$\therefore$ AB = AC . . .[c.p.c.t.]
$\therefore$ $\triangle$ABC is an isosceles triangle.

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Question 62 Marks

$\triangle$ABC, AD is perpendicular bisector of BC. Show that $\triangle$ABC is an isosceles triangle in which AB = AC.

Answer

Given : AD ⊥ BC.
To prove : ABC is an isosceles triangle in which AB = AC.
Proof : In $\triangle$ADB and $\triangle$ADC,
DB = DC . . . [As AD ⊥ bisector of BC]
∠ADB = ∠ADC . . . [Each 90^o]
AD = AD . . .[Common]
$\therefore \triangle \mathrm { ADB } \cong \triangle \mathrm { ADC }$ . . .[By SAS property]
$\therefore$ AB = AC . . .[c.p.c.t]
$\therefore$ $\triangle$ABC is an isosceles triangle in which AB = AC.

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Question 72 Marks

In figure, AC = AE, AB = AD and $\angle$BAD = $\angle$EAC. Show that BC = DE.

Answer

Given : AC = AE, AB = AD and $\angle$BAD = $\angle$EAC.
To prove ; BC = DE
Proof : In DABC and DADE
AC = AE, AB = AD and ∠BAD = ∠EAC ...[Given]
$\therefore$ $\angle$BAD + $\angle$DAC = $\angle$DAC + $\angle$EAC ...[Adding $\angle$DAC to both sides]
$\therefore$ $\angle$BAC = $\angle$DAE ...(1)
AC = AE ...[Given]
$\angle$BAC = $\angle$DAE ...[From (1)]
AB = AD ...[Given]
$\therefore$ DABC $\cong$ DADE ...[By SAS property]
$\therefore$ BC = DE ...[c.p.c.t.]

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Question 82 Marks

ABCD is a quadrilateral in which AD = BC and $\angle$DAB = $\angle$CBA : Prove that:

$\triangle$ ABD $\cong$ $\triangle$ BAC
BD = AC
$\angle$ ABD= $\angle$ BAC

Answer

In quadrilateral ACBD, we have AD = BC and ∠ DAB = ∠ CBA

In ∆ ABC and ∆ BAC,
AD = BC (Given)
∠DAB = ∠CBA (Given)
AB = AB (Common)
$\triangle$ABD $\cong$ $\triangle$BAC ...[By SAS Congruence]
Since ∆ABD ≅ ∆BAC
⇒ BD = AC [By C.P.C.T.]
Since ∆ABD ≅ ∆BAC
⇒ ∠ABD = ∠BAC [By C.P.C.T.]

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Question 92 Marks

In quadrilateral ABCD (See figure). AC = AD and AB bisects $\angle $A. Show that $\triangle$ABC $ \cong $ $\triangle$ABD. What can you say about BC and BD?

Answer

Given: In quadrilateral ABCD, AC = AD and AB bisects $\angle $A.
To prove: $\angle $ABC $ \cong$ $\triangle$ABD
Proof: In $\triangle$ABC and $\triangle$ABD,
AC = AD [Given]
$\angle $BAC = $\angle $BAD [$\because $ AB bisects $\angle $A]
AB = AB [Common]
$\therefore $ $\triangle$ABC $ \cong$ $\triangle$ABD [By SAS congruency]
Thus BC = BD [By C.P.C.T.]

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Question 102 Marks

P is a point equidistant from two lines l and m intersecting at a point A (see figure). Show that AP bisects the angle between them.

Answer

Given: P is a point equidistant from two lines l and m intersect at a point A.
To Prove: AP bisects the angle between them. i.e. $\angle$PAB = $\angle$PAC
Proof: Let PB and PC be perpendiculars from P on lines l and m respectively. Since P is equidistant from lines l and m.
Therefore,PB = PC
In $\Delta$ PAB and $\Delta$PAC, we have
PB = PC [Given]
$\angle$PBA = $\angle$PCA [Each equal to 90^o]
and, PA = PA [Common]
$\triangle$PAB $\cong$ $\triangle$PAC [By RHS congruence criterion]
$\Rightarrow$ $\angle$PAB = $\angle$PAC
Hence Proved

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Question 112 Marks

AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See Figure). Show that the line PQ is the perpendicular bisector of AB.

Answer

In $\triangle$PAQ and $\triangle$PBQ,
AP = BP (Given)
AQ = BQ (Given)
PQ = PQ (Common)
So, $\triangle$PAQ $\cong$ $\triangle$PBQ (SSS rule)
Therefore, $\angle$APQ = $\angle$BPQ (CPCT).
Now let us consider $\triangle$PAC and $\triangle$PBC.
You have: AP = BP (Given)
$\angle$APC = $\angle$BPC ($\angle$APQ = $\angle$BPQ proved above)
PC = PC (Common)
So, $\triangle$PAC $\cong$ $\triangle$PBC (SAS rule)
Therefore, AC = BC (CPCT) .......(i)
$\angle$ACP = $\angle$BCP (CPCT)
and $\angle$ACP + $\angle$BCP = 180^o (Linear pair)
So, $2\angle \mathrm{ACP}$= 180^o
Or, $\angle$ACP = 90^o ......(ii)
From (i) and (ii), we can easily conclude that PQ is the perpendicular bisector of AB.

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Question 122 Marks

AB is a line segment and line l is its perpendicular bisector. If a point P lies on l, show that P is equidistant from A and B.

Answer

Let C be the mid-point of AB.
Clearly, line l passes through C is perpendicular to AB.
In $\triangle $PCA and $\triangle$PCB, we have

AC = BC [$\because$ C is the mid-point of AB]
PC = PC [common side]
$\angle P C A=\angle P C B$ [Each equal to 90^o as l $\perp$ AB]
So, by SAS congruence rule, we obtain
$\triangle P C A \cong \triangle P C B$
$\Rightarrow$ PA = PB

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Question 132 Marks

In Fig., OA = OB and OD = OC. Show that

$\triangle$ AOD $\cong$ $\triangle$ BOC
AD || BC

Answer

You may observe that in $\triangle$ AOD and $\triangle$ BOC,
OA = OB (Given)
OD = OC
Also, since $\angle$ AOD and $\angle$ BOC form a pair of vertically opposite angles,
we have $\angle$ AOD = $\angle$ BOC
So, $\triangle$ AOD $\cong$ $\triangle$ BOC (by the SAS congruence rule)
In congruent triangles AOD and BOC, the other corresponding parts are also equal.
So, $\angle$ OAD = $\angle$ OBC and these form a pair of alternate angles for line segments AD and BC.
Therefore, AD || BC

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