Exercise 5.1: Congruence of Triangles

Question 1:

In the adjoining figure, PA \(\perp\) AB, QB \(\perp\) AB and PA = QB. If PQ intersects AB at O, show that O id the mid-point of AB as well as that of PQ.

Solution:

Given: PA \(\perp\) AB, QB \(\perp\) AB, and PA = QB

To prove: AO = OB and PO = OQ

Proof: In \(\Delta\)APO and \(\Delta\)BPO,

\(\angle\)PAO = \(\angle\)QBO = 90o [Given]

PA = QB [Given]

\(\angle\)PAO = \(\angle\)QBO [Since PA \(\perp\) AB, and QB \(\perp\) AB, PA || QB, and thus PQ is a transversal, therefore, alternate angles are equal]

So, by Angle-Side-Angle criterion of congruence, we have

\(\Delta\)APO \(\cong\) \(\Delta\)BPO

\(\Rightarrow\) AO = OB and PO = OQ [Since corresponding parts of congruent triangles are equal]

Thus, we have O is the midpoint of AB and PQ.

Question 2:

Let the line segments AB and CD intersect at O in such a way that OA = OD and OB = OC. Prove that AC = BD but AC may not be parallel to BD.

Solution:

Given: Line segments AB and CD intersect at O such that OA = OD and OB = OC.

To prove: AC = BD

Proof: In \(\Delta\)AOC and \(\Delta\)BOD, we have

AO = OD [Given]

\(\angle\)AOC  = \(\angle\)BOD [Vertically opposite angles are equal]

OC = OB [Given]

So, by Side-Angle-Side criterion of congruence, we have,

\(\Rightarrow\) AOC \(\cong\) \(\Delta\)BOD

\(\Rightarrow\) AC = BD [Since the corresponding parts of the congruent parts of the congruent triangles are equal]

\(\Rightarrow\) \(\angle\)CAO = \(\angle\)BDO [by c.p.c.t]

Thus, we have, AC = BD

In case   \(\angle\)ODB = \(\angle\)OBD, then \(\angle\)CAO = \(\angle\)OBD which means alternate angles made by lines AB and BD with transversal AB are equal and then lines AC and BD will be parallel.

Question 3:

In the given figure, l||m and M is the mid-point of AB. Prove that M is also the mid-point of any line segment CD having its end points at l and m respectively.

Solution:

Given: Two lines l and m are parallel to each other. M is the midpoint of segment AB. The line segment CD meets AB at M.

To prove: M is the midpoint of CD, that is, CM = MD

Proof:

In \(\Delta\)AMC and \(\Delta\)BMD, we have

\(\angle\)MAC = \(\angle\)MBD [Since l and m are parallel, AB is the transversal, and thus, alternate angles are equal]

AM = MB [Given]

\(\angle\)AMC = \(\angle\)BMD [vertically opposite angles are equal]

So, by Angle-Side-Angle criterion of congruence, we have

\(\Delta\)AMC \(\cong\) \(\Delta\)BMD

Therfore, by corresponding parts of the congruent triangles are equal, we have, CM = MD

Question 4:

In the given figure, AB = AC and OB = OC. Prove that \(\angle\)ABO = \(\angle\)ACO

Solution:

Given: AB = AC and O is an interior point of the triangle such that OB = OC

To prove: \(\angle\)ABO = \(\angle\)ACO

Construction: Join AO

Proof:

In \(\Delta\)AOB and \(\Delta\)AOC, we have

AB = AC [Given]

AO = AO [Common]

OB = OC [Given]

So, by Side-Side-Side criterion of congruence, we have,

\(\Delta\)ABO \(\cong\) \(\Delta\)ACO

\(\Rightarrow\) \(\angle\)ABO = \(\angle\)ACO [by corresponding parts of congruent triangles are equal]

Question 5:

In the given figure, ABC is a triangle in which AB = AC and D is a point AB. Through D, a line DE is drawn parallel to BC and meeting AC at E. Prove that AD = AE.

https://lh6.googleusercontent.com/qZELD14C8fXUP79gidioTQS4OVVCGX8apPCYzcCiFdhCgaHJRRyDhfgsXifaXeZPMIfbg-W-GjAnRxTiO89O0B1YTI1YvSDKsH3t-SV8OnHRWVxgSXirJLjvTtR2sYV_vX7XrYr7Hbj9nXenVQ

Solution:

Given: A \(\Delta\)ABC in which;

AB = AC

and, DE|| BC

To prove: AD = AE

Proof:

Since DE || BC and AB is atransversal.

So, \(\angle\)ADE = \(\angle\)ABC —— (i) [These are corresponding angles]

Also DE || BC and AC is a transversal

So, \(\angle\)AED = \(\angle\)ACB —— (ii) [These are corresponding angles]

But, AB = AC [Given]

So, \(\angle\)ABC = ACB ——- (iii)

as opposite angles are also equal in case sides are equal

So from (i), (ii) and (iii) we have

\(\angle\)ADE = \(\angle\)AED

and in \(\Delta\)ADE, this implies that AD = AE.

Question 6:

In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of \(\Delta\)ABC such that AX = AY. Prove that CX = BY.

https://lh4.googleusercontent.com/1dVWnFppRQNJ12eIgR34Ly8pJuqoGAz0J8fMG2We1XWvkuG5XqyRbHcNH9kdC2Npv5KlDxFrBJRBv7rGPM5znVFNIDcxUWqde1ToNuCkoHUtmza8x0xBS6TLk0t3zpXT88B1q1pwjnqGR7quaQ

Solution:

Given: AX = AY

To prove: CX = BY

Proof: In \(\Delta\)AXC and \(\Delta\)AYB, we have

AX = AY [Given]

\(\angle\)A = \(\angle\)A [Common angle]

AC = AB [Two sides are equal]

So, by Side-Angle-Side criterion of congruence, we have

\(\Delta\)AXC \(\cong\) \(\Delta\)AYB

\(\Rightarrow\) XC = YB [Since corresponding parts of congruent triangles are equal]

Question 7:

In the given figure, C is the mid-point of AB. If \(\angle\)DCA = \(\angle\)ECB and \(\angle\)DBC = \(\angle\)EAC, prove that DC = EC.

https://lh4.googleusercontent.com/EQsAZZkLrIfHmYmu_bvvj18tW7w5De-I97bUnYKlu3_k7rFti7U131Kcaf1bw0961fAl6F5EWZTde4onnQxYedMyvb9kjCzj7RfGP_57vtP6BKBQGNtr45nlSlpM4oQ8QmMiKznbGeqhQ83h0A

Solution:

Given: C is the mid-point of a line segment AB, and D is point such that,

\(\angle\)DCA = \(\angle\)ECB

and \(\angle\)DBC = \(\angle\)EAC

To prove: DC = EC

Proof:

In \(\Delta\)ACE and \(\Delta\)DCB we have;

AC = BC [Given]

\(\angle\)EAC = \(\angle\)DBC [Given]

Also, \(\angle\)DCA = \(\angle\)CDB + \(\angle\)DBA because exterior \(\angle\)DCA in \(\Delta\)DCB is equal to sum of interior opposite angles.

Again in \(\angle\)ACE, we have

ext. \(\angle\)BCE = \(\angle\)CAE + \(\angle\)AEC

But, \(\angle\)DCA = \(\angle\)BCE [Given]

\(\Rightarrow\) \(\angle\)CDB + \(\angle\)DBA = \(\angle\)CAE + \(\angle\)AEC

\(\Rightarrow\) \(\angle\)CDB = \(\angle\)AEC [\(\angle\)DBA = \(\angle\)CAE (Given)]

Thus in \(\Delta\)ACE and \(\Delta\)DCB,

\(\angle\)EAC = \(\angle\)DBC

AC = BC

and, \(\angle\)AEC = \(\angle\)CDB

Thus by Angle-Side-Angle criterion of congruence, we have

\(\Delta\)ACE \(\cong\) \(\Delta\)DCB

The corresponding parts of the congruent triangles are equal.

So, DC = CE [by c.c.p.c.t]

Question 8:

In the given figure, BA \(\perp\) AC and DE \(\perp\) EF such that BA = DE and BF = DC. Prove that AC = EF.

https://lh6.googleusercontent.com/2Ut4RuSCxaCTSfRPhyJQ-1AE2WxMCl_tOyT1WNPqxK5ddCc_kKBG415RXXPDcybxd28t6Mw-RW3-QJKqkwpd27IbgbbNOJLwC9MnjPnvku9m-8QEm7lTO99-Sg7Y9EwHqns_WEudzFvfEvPe2w

Solution:

Given: AB \(\perp\) AC and DE \(\perp\) FE such that,

AB = DE and BF = CD

To prove: AC = EF

Proof:

In \(\Delta\)ABC, we have,

BC = BF + FC

and, in \(\Delta\)DEF

FD = FC + CD

But, BF = CD [Given]

So, BC = BF + FC

and, FD = FC + BF

\(\Rightarrow\) BC = FD

So, in \(\Delta\)ABC and \(\Delta\)DEF, we have,

\(\angle\)BAC = \(\angle\)DEF = 90o [Given]

BC = FD [Proved above]

AB = DE [Given]

Thus, by Right angle-Hypotenuse-Side criterion of congruence, we have

\(\Delta\)ABC \(\cong\) \(\Delta\)DEF

The corresponding parts of the congruent triangle are equal.

So, AC = EF [c.p.c.t]

AE = \(\angle\)BCD [Proved above]

Thus by Angle-Side-Angle criterion of congruence, we have

\(\Delta\)BCD \(\cong\) \(\Delta\)BBAE

The corresponding parts of the congruent triangles are equal.

So, CD = AE [Proved]


Practise This Question

In the plum pudding model which of the following subatomic particles was not included?