Most important question ⁉️

Given:-

ABCD is a parallelogramsuch that the bisectors AP and AP of adjacent sided A and B intersect at P.

To Prove:-

∠APB = 90°

Proof:-

✦ ABCD is a //gm.

∴ AD // BC

⟾ ∠A + ∠B = 180° [ ∵ sum of consecutive interior angle]

⟾ 1/2 ∠A + 1/2 ∠B = 90°

⟾ ∠1 + ∠2 = 90° ---eq. (1)

[ ∵ AP is the bisector of ∠A and BP is the bisector of ∠B]

[ ∴ ∠1 = 1/2 ∠A and ∠2 = 1/2 ∠B ]

✦ Now, in ∆ APB, we have,

⟾ ∠1 + ∠APB + ∠2 = 180° [ ∵ sum of three angles of Δ ]

⟾ 90° + ∠APB + ∠2 = 180° [ ∵ ∠ 1 + ∠2 = 90° from eq.(1) ]

Therefore, ∠APB = 90°

★「 Hence Proved! 」★

ok

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