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In a quadrilateral ABCD, P and R are the midpoints of AB and CD respectively. Also Q and S are points on the sides BC and DA respectively such that BQ = 2QC and DS = 2SA. Prove that the area of the quadrilateral PQRS equals S/2 where S is the area of the quadrilateral ABCD.
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The area of APS is (1/2)*(AD/3)(AB/2)sin( < A).
The area of ABD is (1/2)*AD*ABsin( < A)
So the ratio of the areas of triangles APS and ABD is
(area of APS)/(area of ABD) = {(1/2)*(AD/3)(AB/2)sin( < A)}/{(1/2)*AD*ABsin( < A)} = 1/6
Thus
area of APS = (1/6)*area of ABD ..................(1)
For the same reason, we have
area of CQR = (1/6)*area of BCD ..................(2)
area of DSR = (1/3)*area of ACD ..................(3)
area of BPQ = (1/3)*area of ABC ..................(4)
Adding (1) to (4), we have
area APS + area CQR + area DSR + area BPQ = (1/6)*area of ABD+ (1/6)*area of BCD + (1/3)*area of ACD+(1/3)*area of ABC
= (1/6)*(area of ABD + area of BCD) + (1/3)*(area of BCD + area of ACD)
=S/6 + S/3
=S/2
So area PQRS = S - (area APS + area CQR + area DSR + area BPQ) = S - S/2 = S/2
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