In Fig. 10.40, it is given that RT=TS, 1 =22 and 4 = 23. Prove that Δ RBT Δ SAT.

In the figure, given that:

RT = TS (i)


1 = 22 (ii)


And,


4 = 23 (iii)


To prove:


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


22 = 23 [From (ii) and (iii)]


2 = 3 (iv)


Now, we have in


RT = TS


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


RT = ST [From (i)


TRB = TSA [From (iv)


RTB = STA (Common angle)


By ASA theorem,



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