Q7 of 40 Page 8

AB, CD are two parallel lines and a transversal l intersects AB at G and CD at H. Prove that the bisectors of the interior angles form a parallelogram, with all the angles are right angles.

Given: AB and CD are two parallel lines and the transversal l cut AB at G and CD at H. The bisectors of interior angles intersect at M and N.
To Prove: Quadrilateral NGMH is a rectangle
Proof: AB ïï CD with GH as transversal 
... ∠BGH + ∠DHG = 180°
... ∠ BGH + ∠ DHG = x 180° ⇒ ∠2 + ∠3 = 90°..... (i)
Now,  ∠2 + ∠3 + ∠M = 180°.....(ii) ⇒ 90° + ∠M = 180°
... ∠M = 90°
Similarly ∠N = 90°
Now ∠HGB + ∠HGA = 180°  ......(linear pair)
... ∠HGB + ∠HGA = x 180°
... ∠1 + ∠2 = 90°......(iii)
... ∠NGM = 90°
Similarly ∠NHM = 90°
Since all the four angles of quadrilateral NGMH is 90°
... Quadrilateral NGMH is a parallelogram and rectangle.

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