# 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

2015-02-28T19:12:57+05:30

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

Q+A+1/2 D=180⇒eq 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)
thankq
Ua welcum
2015-02-28T19:14:27+05:30

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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.
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

ADQ + AQD + DAQ = 180
⇒ 0.5 ADC+ Q + A = 180   -----------------(1)
In ΔCBP
CBP + CPB  + BCP = 180
⇒ 0.5 CBA + P + C = 180   ------------------(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)
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THANKS bhaijan .
you are welcome!