Question

in the below figure bisectors of ∠B AND ∠D of a quadrilateral ABCD meets CD and AB , produced to P and Q respectively . prove that ∠P + ∠Q = $\frac{1}{2}$(∠ABC + ∠ADC) . please solve this

Answer 1

In ΔPBC we have,
angle p +pbc+c=180
p+1/2 B+C=180⇒eq 1

In Δqad we have

Q+A+ADQ=180
Q+A+1/2 D=180⇒eq 2

Adding 1 and 2
P+Q+A+C+1/2 B+1/2 A=360

But,A+B+C+D=360
P+Q+A+C+1/2(B +D)=A+B+C+D

P+Q=1/2 (B+D)
P+Q=1/2(ABC+ADC)

Answer 2

BP is the angle bisector of angle ABC,
So angle ABP = angle CBP
DQ is the angle bisector of angle CDA,
So angle CDQ = angle ADQ

In ΔADQ,
ADQ + AQD + DAQ = 180
⇒ 0.5 ADC+ Q + A = 180   -----------------(1)
In ΔCBP
CBP + CPB  + BCP = 180
⇒ 0.5 CBA + P + C = 180   ------------------(2)

adding (1) and (2);
 0.5 ADC+ Q + A + 0.5 CBA + P + C = 180+180
⇒ C + A + 0.5 ABC + 0.5 ADC + P + Q = 360  ---------------(3)

we know that sum of all angles of a quadrilateral = 360
or A+C+ ABC + ADC = 360   ----------------------(4)

from (3) and (4)
A+C+ ABC + ADC = C + A + 0.5 ABC + 0.5 ADC + P + Q
⇒ ABC + ADC = 0.5 ABC + 0.5 ADC + P + Q
⇒ P + Q = ABC - 0.5 ABC + ADC - 0.5 ADC
⇒ P+Q = 0.5 ABC + 0.5 ADC
⇒ P+Q = 0.5(ABC + ADC)