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Given that RT = TS, ∠1 = 2, ∠2 and 4 = 2 ∠(3) Prove that ΔRBT ≅ ΔSAT.
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
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.
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
BD and CE are bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC. Prove that BD = CE
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
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
AO = OB and DO = OC [Corresponding parts of congruent triangles are equal)
Lines AB and CD bisect at O.
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Chapter 10: Congruent Triangles Exercise –...