In Fig., it is given that RT=TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove

1.	In Fig., it is given that RT=TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that Δ RBT ≅ Δ SAT.
Answer» 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

In Fig., it is given that RT=TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that Δ RBT ≅ Δ SAT.

Answer»

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