Question

If two parallel lines are interested by a transversal prove that the bisectors of the angle from a rectangle

Answer 1

$\textbf{ \huge{ Hello Mate! }}$

Given : Two lines AB and CD are ||.

To prove : EFGH is rectangle.

Proof : < AEG = < EGD ( Alternate Angle )

1/2 < AEG = 1/2 < EGD

< FEG = < EGH

Since alternate angles are equal so EF || GH.

Similarily, < EGF = < GEH and again FG || EH.

Hence, EFGH is a ||gm.

< BEG + < EGD = 180° ( Interior consecutive sum )

1/2 ( < BEG + < EGD ) = 1/2 × 180°

< HEG + < EGH = 90°

In triangle EHG.

< EHG + < HEG + < EGH = 180° ( Angle sum pro. )

< EHG + 90° = 180°

< EHG = 90°

Thus, a parallelogram with one angle 90° will be a ractangle.

$\textbf{ \large{ Q.E.D }}$

$\textbf{ Have great future ahead! }$