# 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

2
by dansi902 28.02.2015

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(∠ABC + ∠ADC) . please solve this

2
by dansi902 28.02.2015

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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)

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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)