In a parallelogram ABCD, p is midpoint on the side AB. A line AQ drawn parallel to PC, intersects DC at Q which is midpoint of CD. BD intersects AQ at S and PC at R. Prove that ar $(Δ P R B) = a r (Δ D S Q)$

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Solution

In $Δ P R B a n d Δ D S Q$
Q and P be the mid-points of the parallelogram,
So DQ=PB
And AQ||PC(given)
$∠ P R B = ∠ C R S ($ (vertically opposite angle)
$∠ C R S = ∠ Q S D$ (corresponding angles)
So $∠ P R B = ∠ Q S D$
And $∠ P B R = ∠ Q D S$
SO $Δ P R B$ is congruent to $Δ D S Q$ (By AAS)
so $A r (P R B) = A r (D S Q)$ (Hence proved)

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In a parallelogram ABCD, p is midpoint on the side AB. A line AQ drawn parallel to PC, intersects DC at Q which is midpoint of CD. BD intersects AQ at S and PC at R. Prove that ar (ΔPRB)=ar(ΔDSQ)

In ΔPRB and ΔDSQ Q and P be the mid-points of the parallelogram, So DQ=PB And AQ||PC(given) ∠PRB=∠CRS( (vertically opposite angle) ∠CRS=∠QSD (corresponding angles) So ∠PRB=∠QSD And ∠PBR=∠QDS SO ΔPRB is congruent to ΔDSQ (By AAS) so Ar(PRB)=Ar(DSQ) (Hence proved)

In a parallelogram ABCD, p is midpoint on the side AB. A line AQ drawn parallel to PC, intersects DC at Q which is midpoint of CD. BD intersects AQ at S and PC at R. Prove that ar $(Δ P R B) = a r (Δ D S Q)$