Given that RT $=T S, \angle 1=2, \angle 2$ and $4=2 \angle(3)$ Prove that $\triangle R B T \cong \triangle S A T$.
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