Prove that in a parallelogram, the bisectors of any two consecutive angles intersect at right angles.

Explanation

GIVEN A ||gm ABCD in which the bisectors of two consecutive angles `angle A and angle B ` intersect at a point P.
TO PROVE `angle APB = 90^(@).`
PROOF AD ||BC and AB is a transversal ` " " [ because ABCD " is a ||gm"].`
`therefore angle A + angle B = 180^(@)`
`rArr (1)/(2) angle A + (1)/(2) angle B = 90^(@)`
`rArr angle 1 + angle 2 = 90^(@) " " [ because " AP and BP are bisectors of " angle A and angle B " respectively"]` .
In `triangle APB,` we have
`angle 1 + angle 2 + angle APB = 180^(@) " " ("sum of " angle " of a " triangle)`
`rArr 90^(@) + angle APB=180^(@) rArr angle APB = 90^(@).`