# 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 = (∠ABC + ∠ADC) . please solve this

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by dansi902

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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)

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)

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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)

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)