Question

ABCD is a parallelogram. If angle bi sector of angle A and angle B meet at P. Prove that angle APB = 180 - 1/2 ( angle A+angle B) ​

Answer 1

$\huge{\underline{\underline{\sf{Proof}}}}$

In Δ APB:

∠APB+ ∠BAP+ ∠PBA = 180°

i.e

$∠APB + \frac{∠A}{2} + \frac{∠B}{2} = 180$

So

$∠APB = 180 - ( \frac{∠A}{2} + \frac{∠B}{2} ) - - (1)$

we know that :

$∠A + ∠B + ∠C + ∠D = 360$

(SUM OF THE ANGLES OF QUADRILATERAL)

So

$\frac{∠A}{2} + \frac{∠B}{2} + \frac{∠C}{2} + \frac{∠D}{2} = 180$

$\frac{1}{2} (\frac{∠C}{2} + \frac{∠D}{2}) = 180 - (\frac{∠A}{2} + \frac{∠B}{2}) - - (2)$

Look at eq (1) and eq (2)

we get

$∠APB = 180°- \frac{1}{2} (\frac{∠C}{2} + \frac{∠D}{2})$

Hence Proved