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In ΔRTS, we have
RT=ST
⇒∠TSR=∠TRS (i)
We have,
∠1=∠4 [Vertically opposite angles]
⇒2∠2=2∠3 [∵∠1=2∠2 and ∠4=2∠3 (given)]
⇒∠2=∠3 (ii)
Subtracting (ii) from (i), we get
∠TRS−∠2=∠TSR−∠3
⇒∠TRB=∠TSA (iii)
Thus, in triangles, RBT and SAT, we have
∠RTB=∠STA [Common]
RT=ST [Given]
and, ∠TRB=∠TSA [From (iii)]
So, by ASA congruence criterion, we obtain
ΔRBT≅ΔSAT
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