Question

in the figure bisectors of Angle B and angle D of a quadrilateral ABCD meet CD and ab produced to P and Q respectively prove that angle b + angle Q equals to half of angle abc + angle ADC​

Answer 1

Answer:

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)

Answer 2

Answer:

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