Question

in a parallelogram ABCD, the bisectors of the consecutive angles A and B intersect at p, show that angle APB=90degees
Step-by-step ​

Answer 1

Application of property of parallelogram

Answer: Prove that APB = 90°

Explanation:

Given that in a parallelogram ABCD , bisectors of consecutive angles ∠A and ∠B intersects at P. Need to show that ∠APB = 90°.

Here we will be using property of parallelogram which says that consecutive angles of  parallelogram are supplementary.

=> ∠A + ∠B = 180°

On dividing both the sides by 2 we get

( ∠A + ∠B ) / 2 = 180 / 2

=> 1/2 ∠A + 1/2∠B = 90°    --------eq(1)

Now consider Δ APB

∠PAB + ∠PBA + ∠APB = 180°      [ Angle Sum property of Δ ]

As AP is bisector of ∠A , it means  ∠PAB = 1/2∠A and

And As BP is bisector of ∠B , it means ∠PBA = 1/2∠B

=> (1/2)∠A+ (1/2) ∠B + ∠APB = 180°

From eq(1) (1/2)∠A+ (1/2) ∠B  = 90° ,

=> 90° + ∠APB  = 180°

=> ∠APB  = 180° - 90° = 90°

=> ∠APB  = 90°

Coming up the most satisfying word of geometry  "HENCE PROVED".

#answerwithquality

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