In the figure, given that:
RT = TS (i)
∠1 = 2∠2 (ii)
And,
∠4 = 2∠3 (iii)
To prove: ΔRBT ≅ ΔSAT
Let the point of intersection of RB and SA be denoted by O.
Since, RB and SA intersect at O.
∠AOR = ∠BOS (Vertically opposite angle)
∠1 = ∠4
2∠2 = 2∠3 [From (ii) and (iii)]
∠2 = ∠3 (iv)
Now, we have in ΔTRS
RT = TS
ΔTRS is an isosceles triangle
Therefore, ∠TRS = ∠TSI (v)
But, we have
∠TRS = ∠TRB + ∠2 (vi)
∠TSR = ∠TSA + ∠3 (vii)
Putting (vi) and (vii) in (v), we get
∠TRB + ∠2 = ∠TSA + ∠3
∠TRB = ∠TSA [From (iv)]
Now, in ΔRBT and ΔSAT
RT = ST From (i)
∠TRB = ∠TSA From (iv)
∠RTB = ∠STA (Common angle)
By ASA theorem,
ΔRBT ≅ ΔSAT
