CBSE IX Congruence of Triangle Solved Questions

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.

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)

Q. 2. ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that ∠BCD is a right angle.

In ΔABC,

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 )

AM  =  PN                           ( 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

Q. 6. In triangle PQR, PQ> PR. QS and RS are the bisectors of angle Q and angle R. Prove that SQ> SR

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)]

Now, in ∆SQR, we have    ∠SRQ > ∠SQR       [proved above]

⇒ 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

AB+BC+AC> 2AD

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.

Let us join AC.

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

Let us join BD.

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

Q.10.  If S. is any point on the side QR of triangle PQR, prove that PQ+QR+RP> 2PS

 In   ΔPQS,

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.

Q.11. Prove that the difference of any two sides of a triangle is less than the third side.

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

But, BD = AB – AD and AD = AC         => BD = AB – AC

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.

A similar argument shows that 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

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

BD = BC ; AB = AB and <B = <B = 90⁰

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.

Solution:

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.

Solution:

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

Let BP be the bisector of ∠ABC. Join PD.

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)

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)

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)

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