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