Chapter 10: Congruent Triangles Exercise – 10.2
Question: 1
Given that RT = TS, ∠1 = 2, ∠2 and 4 = 2 ∠(3) Prove that ΔRBT ≅ ΔSAT.
Solution:
In the figure, given that
RT = TS .... (i)
∠1 = 2∠2 .... (ii)
And ∠4 = 2∠3 .... (iii)
To prove that ΔRBT ≅ ΔSAT.
Let the point of intersection RB and SA be denoted by O
Since RB and SA intersect at O
∠AOR = ∠BOS [Vertically opposite angles]
∠1 = ∠4
2∠2 = 2∠3 [From (ii) and (iii)]
∠2 = ∠3 ….(iv)
Now we have RT = TS in ΔTRS
ΔTRS is an isosceles triangle
∠TRS = ∠TSR .... (v)
But we have
∠TRS = ∠TRB + ∠2 ..... (vi)
∠TSR = ∠TSA + ∠3 .... (vii)
Putting (vi) and (vii) in (v) we get
∠TRB + ∠2 = ∠TSA + ∠B
⟹ ∠TRB = ∠TSA [From (iv)]
Now consider ΔRBT and ΔSAT
RT = ST [From (i)]
∠TRB = ∠TSA [From (iv)] ∠RTB = ∠STA [Common angle]
From ASA criterion of congruence, we have
ΔRBT = ΔSAT
Question: 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.
Such that BC ∥ 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 ΔAOD ≅ ΔBOC
Question: 2
BD and CE are bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC. Prove that BD = CE
Solution:
Given that ΔABC is isosceles with AB = AC and BD and CE are bisectors of ∠B and ∠ C We have to prove BD = CE
Since AB = AC
⟹ ΔABC = ΔACB ... (i)
[Angles opposite to equal sides are equal]
Since BD and CE are bisectors of ∠B and ∠C
∠ABD = ∠DBC = ∠BCE = ECA = ∠B/2 = ∠C/2
Now,
Consider ΔEBC = ΔDCB
∠EBC = ∠DCB [∠B = ∠C] [From (i)]
BC = BC [Common side]
∠BCE = ∠CBD [From (ii)]
So, by ASA congruence criterion, we have Δ EBC ≅ Δ DCB
Now,
CE = BD [Corresponding parts of congruent triangles we equal]
or, BD = CE
Hence proved
Since AD ∥ BC and transversal AB cuts at A and B respectively
∠DAO = ∠OBC .... (ii) [alternate angle]
And similarly AD ∥ BC and transversal DC cuts at D and C respectively
∠ADO = ∠OBC …. (iii) [alternate angle]
Since AB end CD intersect at O.
∠AOD = ∠BOC [Vertically opposite angles]
Now consider ΔAOD and ΔBOD
∠DAO = ∠OBC [From (ii)]
AD = BC [From (i)]
And ∠ADO = ∠OCB [From (iii)]
So, by ASA congruence criterion, we have
ΔAOD ≅ ΔBOC
Now,
AO = OB and DO = OC [Corresponding parts of congruent triangles are equal)
Lines AB and CD bisect at O.
Hence proved