CBSE Exam Congruence
of Triangle Solved Questions

Q. 1. Prove that Sum of Two Sides of a triangle is greater than twice
the length of median drawn to third side.

**Given:**Δ ABC in which AD is a median.

**To prove:**AB + AC > 2AD.

**Construction:**Produce AD to E, such that AD = DE. Join EC.

**Proof:**In ΔADB and ΔEDC,

AD =
DE
(Construction)

BD =
BD (D
is the mid point of BC)

∠ADB = ∠EDC
(Vertically opposite angles)

∴ ΔADB ≅
ΔEDC
(SAS congruence criterion)

⇒ AB = ED
(CPCT)

In
ΔAEC,

AC +
ED > AE (

**Sum of any two sides of a triangles is greater than the third side**)
∴ AC + AB > 2AD
(

**AE = AD + DE = AD + AD = 2AD & ED = AB**)
AB = AC (Given)

⇒ ∠ACB = ∠ABC (Angles
opposite to equal sides of a triangle are also equal)

In ΔACD, AC
= AD

⇒ ∠ADC = ∠ACD (Angles
opposite to equal sides of a triangle are also equal)

In ΔBCD,

∠ABC + ∠BCD + ∠ADC = 180º
(Angle sum property of a triangle)

⇒ ∠ACB + ∠ACB +∠ACD + ∠ACD = 180º

⇒ 2(∠ACB + ∠ACD) = 180º

⇒ 2(∠BCD) = 180º

⇒ ∠BCD = 90º

Q.3.Given:
two triangles ABC and PQR in which AB=PQ, BC=QR , median AM =median PN prove
that triangle ABC is congruent to triangle PQR.

In ∆ ABM
and ∆ PQN

AB = PQ
( Given )

And BM = QN ( As M and N are the midpoint of sides BC and QR respectively and given BC= QR ) ∆ ABM ≅ ∆ PQN ( By SSS rule )

SO, ∠ ABM = ∠ PQN ( by CPCT )

Now In ∆ ABC and ∆ PQR

AB = PQ ( Given )

BC = QR ( Given )

And ∠ ABC = ∠ PQR ( As we proved )

∆ ABC ≅ ∆ PQR ( By SAS rule )
( Hence proved )

Q.4. The vertex angle of an isosceles triangle is twice the sum of its base angles.
Find the measure of all the angles.

Let ABC be an isosceles ∆.Let the measure of each of the base angles =

*x*
Let ∠B = ∠C =

*x*
Now, vertex angle = ∠A = 2

*x*
Now,∠A + ∠B + ∠C = 180° [angle sum property]

⇒2

*x*+*x*+*x*= 180°⇒4*x*= 180⇒*x*= 180/4=45^{0}
So, measure of each of the base angles = 45°

Now, measure of the vertex angle = 90°

Q. 5. Prove that the triangle formed by joining
the midpoints of the sides of an equilateral triangle is also equilateral.

Let DEF be the midpoints of sides of a triangle ABC( with
D on BC, E on AB and F on AC ).

Now, considering
triangles AEF and ABC, angles

EAF = BAC and AE / AB = 1/2 and AF/AC = 1/2.

Hence, both triangles are similar by the SAS ( Side -
Angle - Side ) criterion and correspondingly as AE/AB=AF/AC=EF/BC ( similar
triangle properties ), EF =BC/2.

The cases DF=AC/2 and DE=AB/2 can be proved in the same
way.

So, AB=BC=AC (from the given data)

2DF=2EF=2DE

DE=EF=DF

So triangle DEF is also Equilateral Triangle

The triangle formed by joining the mid-points of the
equilateral triangle is also an equilateral triangle

In ∆PQR, we have,

PQ > PR [given]

⇒ ∠PRQ > ∠PQR [angle opposite to longer side of a ∆ is greater]

⇒12∠PRQ > 12∠PQR ........(1)

Since, SR bisects ∠R, then∠SRQ = 1/2∠PRQ ........(2)

Since SQ bisects ∠P, then∠SQR = 1/2∠PQR .......(3)

Now, from (1), we have 1/2∠PRQ > 1/2∠PQR

⇒∠SRQ > ∠SQR [using (2) and (3)]

⇒ SQ > SR [side opposite to greater angle of a ∆ is longer]

Q.7. In triangle
ABC (A at the top) , D is any point on the side BC. Prove that AB+BC+CA 2AD

In triangle
ABD,

AB+BD >AD
(Sum of two sides of a triangle is greater than the third side) ... (1)

In triangle
ACD,

AC+CD>AD
(Sum of two sides of a triangle is greater than the third side) ...(2)

Adding eq. (1)
and (2)

AB+(BD+CD)+AC>
AD+AD

Q.8. In triangle ABC, if AB
is the greatest side, then prove that angle c is greater than 60 degrees

It is given that, AB is the longest side of the ∆ABC.

AB > BC and AB > AC.Now, AB > BC⇒∠C > ∠A (angle opposite to longer side is greater) ....(1)

Also,AB > AC⇒∠C > ∠B (angle opposite to longer side is greater) ....(2)

adding (1) and (2) ,

we get∠C + ∠C > ∠A + ∠B

⇒2∠C > ∠A + ∠B⇒2∠C + ∠C > ∠A + ∠B + ∠C⇒3∠C > 180°⇒∠C > 60°

Q.9. AB and CD are
respectively the smallest and longest sides of a quadrilateral ABCD (see the
given figure). Show that ∠A > ∠C and ∠B > ∠D.

In Δ ABC,

AB < BC (AB
is the smallest side of quadrilateral ABCD)

∴ ∠2 < ∠1 (Angle opposite to the smaller side is smaller) ...
(1)

In ΔADC,

AD < CD (CD
is the largest side of quadrilateral ABCD)

∴ ∠4 < ∠3 (Angle opposite to the smaller side is smaller) ...
(2)

On adding
equations (1) and (2), we obtain

∠2 + ∠4 < ∠1 + ∠3

⇒ ∠C < ∠A

⇒ ∠A > ∠C

In ΔABD,

AB < AD (AB
is the smallest side of quadrilateral ABCD)

∴ ∠8 < ∠5 (Angle opposite to the smaller side is smaller) ...
(3)

In ΔBDC,

BC < CD (CD
is the largest side of quadrilateral ABCD)

∴ ∠7 < ∠6 (Angle opposite to the smaller side is smaller) ...
(4)

On adding
equations (3) and (4), we obtain

∠8 + ∠7 < ∠5 + ∠6

⇒ ∠D < ∠B ⇒ ∠B > ∠D

PQ + QS >
PS (i) ……………..(Sum of two sides of a triangle is greater than the
third side)

In ΔPSR,

PR + SR >
PS ……(ii)… Sum of two sides of a triangle is greater than the third
side)

Adding (i) and
(ii), we get

PQ + QS + PR +
SR > 2PS

PQ + QR + PR
> 2PS (QS + SR = QR) Hence proved.

Construction: Take a Point D on AB
such that AD = AC and join CD

Prove that : AB – AC < BC , AB – BC
< AC and BC-AC <AB

Proof: In Δ ACD, Ext <4 > <2

but ,
AD = AC => <1 = <2

So , < 4 > < 1 ----------------(i)

Now , In Δ BCD, ext <1 > <3
-------------(ii)

Then from (i) and
(ii)

< 4
> <3 => BC > BD

So, BC > AB – AC

Q.12. that Sum of any two sides of triangle is greater than third side .

Solution:.

Construction:
Extend BA to D Such that AD = AC

Proof
: In Δ DACD,

*DA*=*CA*.
Therefore, ∠

*ADC*=∠*ACD*[ isosceles triangle have two equal angles]
∠

*ADC + <1*> ∠*ACD*
Thus, ∠

*BCD*>∠*BDC*[by Euclid's fifth common notion.]
In △

*DCB*
∠

*BCD*> ∠*BDC*, So,*BD*>*BC*.
But

*BD*=*BA*+*AD*, and*AD*=*AC*.
Thus,

*BA*+*AC*>*BC*.*AC*+

*BC*>

*BA*and

*BA*+

*BC*>

*AC*.

**OR, Another way to prove**

Draw a triangle, △ ABC and
line perpendicular to AC passing through vertex B.

Prove that BA + BC > AC

Prove that BA + BC > AC

From the diagram, AM is the shortest distance from vertex A to BM. and CM is the shortest distance from vertex C to BM.

i.e. AM < BA and CM < BC

By adding these inequalities, we have

AM + CM < BA + BC

=> AC < BA + BC (∵ AM + CM = AC)

BA + BC > AC (Hence Proved)

Q.13. if one acute angle in a
right angled triangle is double the other then prove that the hypotenuse is
double the shortest side

Given: In Δ ABC , <B =
90

^{0}and <ACB = 2 <CAB
Prove that AC = 2BC

Construction: Produce CB to D such that BC = BD Join
to AD

Proof : In Δ ABD, and ABC

^{0}

By SAS congruency , D ABD ≅ D ABC

By CPCT, AD = AC

<DAB =
<BAC = X

^{0}
So, < DAC
= 2X

^{0}
=> <ACB = <ACD

Now in Triangle Δ ADC, <DAC
= <ACD= 2X

^{0}
So, AD = DC

=> AC = DC = 2BC
Proved

Q. 14. Prove that in a triangle the side opposite
to the largest angle is the longest.

Given , in Δ ABC, <ABC < <ACB

There is a triangle ABC, with angle ABC > ACB.

Assume line AB = AC

Then angle ABC = ACB, This is a contradiction

Assume line AB > AC

Then angle ABC < ACB, This also contradiction our
hypothesis

So we are left with only one possibility ,AC> AB,
which must be true

Hence proved: AB
< AC

Q. 15. Prove that in a triangle the angle opposite
to the longer side is the longest.

Given, in Δ ABC, AC
> AB.

Construction: Take a point D on AC such that AB = AD

Proof: Angle ADB > DCB

< ADB = <ABD

So < ABD > <DCB (or ACB)

< ABC >
<ABD, so < ABC > <ACB

Q. 16.In a Δ ABC ,<B =
2<C. D is a point on BXC such that AD bisect < BAC and AB = CD. Prove
that < BAC = 72 degree

In ΔABC, we have

∠B = 2∠C or, ∠B = 2

*y*, where ∠C =*y*
AD is the bisector of ∠BAC. So, let ∠BAD = ∠CAD =

*x*
In ΔBPC, we have

∠CBP = ∠BCP =

*y*⇒ BP = PC ... (1)
Now, in ΔABP and ΔDCP, we have

∠ABP = ∠DCP =

*y*
AB = DC [Given]

and, BP = PC [Using (1)]

So, by SAS congruence criterion, we have

Δ ABP ≅ Δ DCP

<BAP = < CPD and AP = DP

<CDP = 2x
then <ADP = < DAP = x
[<A = 2x]

In ΔABD, we have

∠ADC = ∠ABD + BAD ⇒

*x*+ 2*x*= 2*y*+*x*⇒*x*=*y*
In ΔABC, we have

∠A + ∠B + ∠C = 180°

⇒ 2

*x*+ 2*y*+*y*= 180°
⇒ 5

*x*= 180°
⇒

*x*= 36°
Hence, ∠BAC = 2

*x*= 72°You may also use this way:

Q.17, If
o is any point in the interior of triangle ABC .Prove that

(a) AB + AC
> OB + OC

(b) AB + BC +
CA > OA + OB + OC

(c )OA +OB+OC>1/2(AB+BC+CA)

(c )OA +OB+OC>1/2(AB+BC+CA)

Construction:
Produce BO to meet AC at D

In D ABD, AB
+ AD > BD => AB + AD > OB + OD
------(i)

In D OCD, OD
+ DC > OC ------(ii)

Adding (i)
and (ii) we get,

AB +
AD + OD + DC > OB + OD + OC

=> AB + AC > OB + OC --------- (iii)

**Hence prove (a)**
Similarly we get ,

BC + BA > OA + OC ---------(iv)

and , CA + CB > OA + OB ---------(v)

and , CA + CB > OA + OB ---------(v)

Adding (iii),(iv)and (v) we get,

2(AB + BC + CA) > 2(OA + OB + OC)

AB + BC + CA > OA + OB + OC

**Hence prove (b)**
In D OAB , D OBC and D OCA

[OA + OB > AB ] + [OB + OC>BC] + [ OC + AO > AC]

2[OA + OB + OC] > AB + BC
+ CA

[OA + OB + OC] > ½ [AB + BC + CA]

[OA + OB + OC] > ½ [AB + BC + CA]

**Hence prove (c)**
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