Thread: geometry

Question
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.

Solution
Line P R divide quadrilateral ABCD with equal area s/2:
A (APRD) = A (PBCR) = S / 2

Sum of the triangle area PRS and PQR is area of the quadrilateral PQRS and is s/2.

It is s/2 not only for BQ = 2QC and DS = 2SA, it is true for
BQ = N*QC and DS = N*SA
where N=(1/M or M) and M real number.

Triangle area PRS(1/M) equal PQR(M), sum=S/2 and
PQR(1/M) equal PRS(M), sum=S/2

I was wondering if this is how you do the question?

Originally Posted by xwrathbringerx

Question
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.

Solution
Line P R divide quadrilateral ABCD with equal area s/2:
A (APRD) = A (PBCR) = S / 2
Sum of the triangle area PRS and PQR is area of the quadrilateral PQRS and is s/2.

It is s/2 not only for BQ = 2QC and DS = 2SA, it is true for
BQ = N*QC and DS = N*SA
where N=(1/M or M) and M real number.

Triangle area PRS(1/M) equal PQR(M), sum=S/2 and
PQR(1/M) equal PRS(M), sum=S/2

I was wondering if this is how you do the question?

Nowhere in the posted question is said, nor is it implied, that the line L divides the quadrilateral ABCD into two equal parts.
Although it is said that the L joins the midpoints of AB and CD, that does not mean that L divides ABCD equally such that APRD = PBCR in areas.

To imagine that, think when ABCB is like a "trapezoid" but the shorter upper base, BC, is not parallel to he lower base, AD.

Can anyone help me solve this?!?