CBSE IX Congruence of Triangle Solved Questions
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.
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 = 2x
Now,∠A + ∠B + ∠C = 180° [angle sum property]
⇒2x + x + x = 180°⇒4x = 180⇒x = 180/4=450
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.
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
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjTPnjX4ynEFJMK5MIPMW54zmiTYkqe1JCwLT47znaPyta3nNqqC8u0uUk_RIZD3-Kxl2MlWJLO_thlB26fZom-KPPXQ_pA2aTNa6RAZg17-bkhDg3SuR1RmcDrmEbPQZYEXrJt6y1p7Kc/s1600/congruence+of+triangle-13.png)
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 =
900 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
By SAS congruency , D ABD ≅ D ABC
By CPCT, AD = AC
<DAB =
<BAC = X0
So, < DAC
= 2X0
=> <ACB = <ACD
Now in Triangle Δ ADC, <DAC
= <ACD= 2X0
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 = 2y,
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 +
2x = 2y + x ⇒ x = y
In ΔABC, we have
∠A + ∠B + ∠C = 180°
⇒ 2x +
2y + y = 180°
⇒ 5x =
180°
⇒ x =
36°
Hence, ∠BAC = 2x =
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] Hence prove (c)
[OA + OB + OC] > ½ [AB + BC + CA] Hence prove (c)
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