Question

Pls help 15 points question

Answer 1

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)

3.In triangle PQM

=> PMQ + MPQ + Q = 180 [Angle sum property of a triangle]

=> 90 + MPQ + Q = 180 (PMQ = 90)

=> Q = 90 - MPQ

=> 90 - MPQ = Q _____(1)

In triangle PMR

=> PMR + PRM + R = 180

=> 90 + MPR + R = 180

=> R = 90 - MPR

=> 90 - MPR = R _____(2)

Subtracting (1) and (2), we get

=> (90 - MPQ) - (90 - MPR) = Q - R

=> MPR - MPQ = Q - R

=> (RPA +APM) - (QPA - APM) = Q - R

=> QPA + APM -QPA + APM = Q - R [ As PA is bisector of QPR so, RPA=QPA]

=> 2 APM = Q - R

=> APM = 1/2 (Q - R)

Answer 2

Answer:

refer to the attachment above for ur answer

hope it helps you ^_^