Question

In the figure, it is given $TS=TR,\angle 1=2\angle =$ and $\angle 4=2\angle 2\angle 3$. Prove that $△RBT\cong △SAT$.

Answer 1

Given,

$TS=TR$...(i)

$\angle 1=2\angle 2$...(ii)

$\angle 4=2\sqrt{3}$...(iii)

$\angle 1=\angle 4$ Vertically Opposite Angles

$\therefore 2\angle 2=2\angle 3\Rightarrow \angle 2=\angle 3$...(iv)

$\angle \therefore \angle TRS=\angle TSR$...(A) Isosceles triangle

$\angle TRS=\angle TRB+\angle 2$...(B)

$\angle TSR=\angle TSA+\angle 3$...(C)

substitute (B) and (C) in (A), we get,

$\angle TRB+\angle 2=\angle TSA+\angle B$

$\Rightarrow \angle TRB=\angle TSA$

$RT=ST$...From (i)

$\angle TRB=\angle TSA$...From (iv)

$\angle RTB=\angle STA$

$\therefore △RBT\cong △SAT$ From ASA