Question

If a figure AP and bp are bisectors of angle A and Angle B of parallelogram ABCD then prove that the angle apb=90

Answer 1

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! 」★

________________

Answer 2

Answer:

If AP and BP are angle bisector, then

∠1=∠2=∠A/2

∠3=∠4=∠B/2

And Consecutive angles are supplementary

∴∠1+∠2=180−(∠3+∠4)

∠A=180−∠B

From the figure, ∠APB=180−(∠1+∠3)

=180−(180−(∠2+∠4))

=∠2+∠4=∠A/2+∠B/2

⇒2∠APB=∠A+∠B

thus APB=90