Question

Prove that sum of all angles of a triangle is ${180}^{\circ }$.

Answer 1

$ST||BC$

then $\angle SAB=\angle B$ (Altemate interior angles)

also $\angle TAC=\angle C$ (Altemate interior angles)

Then at point $A$

$\angle SAB+\angle A+\angle TAC={180}^o$ (Linear pair)

$\angle B+\angle A+\angle C={180}^o$

$\angle A+\angle B+\angle C={180}^o$