Question

in the given figure angle Q greater than angle R,PA is the bisector of angle QPR and PM perpendicular QR .Prove that​

Answer 1

Answer:

In△PAQ

Sumoftwooppositeinteriorangles=Exteriorangle

α=θ+∠θ−(i)

In△PAR

α+θ+∠R=180∘

α=180∘−θ−∠R−(ii)

In△PMQ

∠MPQ=180−∠Q−90∘

=90−∠Q

∠APM=∠APQ−∠MPQ

=θ−(90−∠Q)

=θ+∠Q−90∘−(iii)

∵(i)=(ii)

=>θ+∠Q=180−θ−∠R

=>2θ=180−∠Q−∠R

=>θ=90−21(∠Q+∠R)

puttingthisvalueofθin(iii)

∠APM=∠Q+90−21(∠Q

Step-by-step explanation:

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