Question

AP and BP are the bisectors of two adjacent angles A and B of quadrilateral ABCD.Prove that 2angle APB=angle C+ angle D.

Answer 1

A+B+C+D= 360°

C+D= 360°-A-B

C+D= 360°- 2A/2- 2B/2

Therefore after rearranging we get,

C+D= (180- A/2- B/2) + (180- A/2- B/2)

C+D= 2AOB

Since, (180- A/2- B/2)= AOB

Multiplying L.H.S. & R.H.S. by m we get,

m.C+ m.D= 2.m.AOB

2.m.AOB= k.m.AOB

Therefore, k=2

Answer 2

Given a quadrilateral ABCD with AP and BP are bisectors of two adjacent angles A and B. Prove that $2\angle APB=\angle C+\angle D$.

Explanation:  

• In the given quadrilateral ABCD the angle bisectors AP and BP of angles A and B form two triangles APD and BPC.
• Now we know that the sum of all angles of a triangle is $180$° and a quadrilateral is $360$°.
• Therefore, $\angle A+\angle B+\angle C+\angle D=360$ --(a)
• Hence in triangle APD we get, $\angle\frac{A}{2}+\angle D+ \angle DPA=180$  --(b)
• Similarly in triangle BPC we get, $\angle\frac{B}{2}+\angle C+ \angle BPC=180$  --(c)
• From (a) and Adding (b) to (c) we get, $\angle\frac{A+B}{2}+\angle D+\angle C+ \angle DPA+\angle BPC=360 \\->\angle A+\angle B+2\angle D+2\angle C+ 2(\angle DPA+\angle BPC)=720\\->(\angle A+\angle B+\angle D+\angle C)+\angle D+\angle C+2(\angle DPA+\angle BPC)=720\\->\angle D+\angle C+2(\angle DPA+\angle BPC)=360$
• Now we know that at point P, $\angle DPA+\angle BPC+\angle APB=180$  
• Hence we get,
• $->\angle C+\angle D+2(180-\angle APB)=360\\->\angle C+\angle D+360-2\angle APB=360\\->2\angle APB=\angle C+\angle D$   --->Hence proved.