Question

In Fig. 10.40, it is given that RT = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that ΔRBT ≅ ΔSAT. 

 

Answer 1

In the figure given that 

RT=TS                     .......................(1)

∠=2 ∠2                   ........................(2) 

And `∠4=2sqrt3`     ......................(3) 

And given 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 angles] 

⇒ ∠1=∠4 

⇒2∠2=2∠3                           [From (2) and (3)] 

⇒ ∠2=∠3                             ......................(4) 

Now we have RT= TS  in ΔTRS 

⇒ΔTRS  is an isosceles triangle 

∴∠TRS=∠TSR                 ............................(5) [Angles opposite to equal sides are equal] 

But we have 

∠TRS=∠TRB+∠2              ....................(6)

And ∠TSR=∠TSR+∠3      .....................(7) 

Putting (6) and (7) in (5) we get  

∠TRB+∠2=∠TSA+∠B 

⇒ ∠TRB=∠TSA                    [∵ from (4)]

Now consider ΔRBT and ΔSAT 

RT =ST                     [From (1)] 

∠TRB=∠TSA               [From (4)] 

∠RTB=∠STA              [Common angle] 

From ASA criterion of congruence, we have ΔRBT≅ΔSAT