Question

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

Answer 1

Solution:-

Consider $△PQR$ with angles $\angle 1,\angle 2$ and $\angle 3$

Draw a line XY passing through P parallel to QR

For lines XY and QR with transversal PQ

$\angle 2=\angle 4$ (alternate angles)

For lines XY and QR with transversal PR

$\angle 3=\angle 4$ (Alternate angles)

Also, for line XY

$\angle 1+\angle 4+\angle 5={180}^{\circ }$ (linear pair)

$\angle 1+\angle 2+\angle 3={180}^{\circ }$

Hence, sum of all angles of a triangle is ${180}^{\circ }$