RD Sharma Solutions Class 9 Congruent Triangles Exercise 10.2

RD Sharma Class 9 Solutions Chapter 10 Ex 10.2 Free Download

RD Sharma Solutions Class 9 Chapter 10 Ex 10.2

(1) In fig. (10).40, it is given that RT = TS, \(\angle\) 1 = 2 \(\angle\) 2 and 4 = 2 \(\angle\) (3) Prove that \(\Delta RBT\) \(\cong \Delta SAT\).

Solution:

In the figure, given that

RT = TS                                                                          ……(i)

\(\angle\) 1 = 2 \(\angle\) 2            ……(ii)

And \(\angle\) 4 = 2 \(\angle\) 3    ……(iii)

To prove that \(\Delta RBT\) \(\cong \Delta SAT\).

Let the point of intersection RB and SA be denoted by O

Since RB and SA intersect at O

\(\angle\) AOR = \(\angle\) BOS          [Vertically opposite angles]

  • \(\angle\) 1 = \(\angle\) 4
  • 2 \(\angle\) 2 = 2 \(\angle\) 3 [From (ii) and (iii)]
  • \(\angle\) 2 = \(\angle\) 3 ……(iv)

Now we have RT =TS in \(\Delta\) TRS

\(\Delta\) TRS is an isosceles triangle

\(\angle\) TRS = \(\angle\) TSR              ……(v)

But we have

\(\angle\) TRS = \(\angle\) TRB + \(\angle\) 2                                                              ……(vi)

\(\angle\) TSR = \(\angle\) TSA + \(\angle\) 3                                                              ……(vii)

Putting (vi) and (vii) in (v) we get

\(\angle\) TRB + \(\angle\) 2 =  \(\angle\) TSA + \(\angle\) B

=> \(\angle\) TRB = \(\angle\) TSA        [From (iv)]

Now consider \(\Delta\) RBT and \(\Delta\) SAT

RT = ST [From (i)]

\(\angle\) TRB = \(\angle\) TSA            [From (iv)]  \(\angle\) RTB = \(\angle\) STA    [Common angle]

From ASA criterion of congruence, we have

\(\Delta\) RBT = \(\Delta\) SAT

 

(2) Two lines AB and CD intersect at O such that BC is equal and parallel to AD. Prove that the lines AB and CD bisect at O.

Solution: Given that lines AB and CD Intersect at O

 8

Such that BC \(\parallel\) AD and  BC = AD  …….(i)

We have to prove that AB and CD bisect at O.

To prove this first we have to prove that \(\Delta\) AOD \(\cong\) \(\Delta\) BOC

 

(3) BD and CE are bisectors of \(\angle\) B and \(\angle\) C of an isosceles \(\Delta\) ABC with AB = AC. Prove that BD = CE

Solution:

Given that \(\Delta\) ABC is isosceles with AB = AC and BD and CE are bisectors of \(\angle\) B and \(\angle\) C We have to prove BD = CE

Since AB = AC

=> \(\Delta\) ABC = \(\Delta\) ACB ……(i)

[Angles opposite to equal sides are equal]

Since BD and CE are bisectors of \(\angle\) B and \(\angle\) C

  • \(\angle\) ABD = \(\angle\) DBC = \(\angle\) BCE = ECA = \(\frac{\angle B}{2}\) = \(\frac{\angle C}{2}\)

Now,

Consider \(\Delta\) EBC = \(\Delta\) DCB

9

\(\angle\) EBC = \(\angle\) DCB [\(\angle\) B = \(\angle\) C] [From (i)]

BC = BC [Common side]

\(\angle\) BCE = \(\angle\) CBD [From (ii)]

So, by ASA congruence criterion, we have \(\Delta\) EBC \(\cong\) \(\Delta\) DCB

Now,

CE = BD     [Corresponding parts of congruent triangles we equal]

or, BD = CE

Hence proved

Since AD \(\parallel\) BC and transversal AB cuts at A and B respectively

\(\angle\) DAO = \(\angle\) OBC  …….(ii) [alternate angle]

And similarly AD \(\parallel\) BC and transversal DC cuts at D and C respectively

\(\angle\) ADO = \(\angle\) OBC                 ..…(iii) [alternate angle]

Since AB end CD intersect at O.

\(\angle\) AOD = \(\angle\) BOC               [Vertically opposite angles]

Now consider \(\Delta\) AOD and \(\Delta\) BOD

\(\angle\) DAO = \(\angle\) OBC [From (ii)]

AD = BC                                                                        [From (i)]

And \(\angle\) ADO = \(\angle\) OCB  [From (iii)]

So, by ASA congruence criterion, we have

\(\Delta AOD \cong \Delta BOC\)

Now,

AO= OB and DO = OC [Corresponding parts of congruent triangles are equal)

  • Lines AB and CD bisect at O.

Hence proved

Leave a Comment

Your email address will not be published. Required fields are marked *