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 to half of the area of ABCD?

Answer 1

Indeed `S_(PQRS)=S_(ABCD)` as explained.

Explanation:

There's a simple way to solve this problem using Vectorial Product (in case cross product).

The formula for a quadrilateral's area using vectors is discussed in

Quadrilateral's area using vectors

So the area of a quadrilateral is given by
`S=(1/2)*|vec p xx vec q|`, where `vec p` and `vec q` are the vectors of the diagonals of the quadrilateral

(Let's remind that `vec a xx vec a=0` and `vec a xx vec b=-vec b xx vec a`)

In case:
`S_(ABCD)=(1/2)*|AvecC xx BvecD|`
`S_(PQRS)=(1/2)*|PvecR xx SvecQ|`

To be able to compare the results we should get the vectors as function of other vectors in common. In this problem, it will be convenient to get the vectors of the diagonals as function of `A vec B` (or its parts such as `A vec P` or `P vec B`), `P vec R` and `D vec C` (or its parts such as `D vec R` or `R vec C`).

Quadrilateral ABCD
`A vec B=AvecP+PvecR+RvecC=AvecP+PvecR+DvecR`
`BvecD=-PvecB+PvecR-DvecR=-AvecP+PvecR-DvecR`
So
`S_(ABCD)=(1/2)| (A vec P+P vec R+D vec R) xx (-A vec P+P vec R-D vec R) |`
`=(1/2)*|AvecP xx PvecR-cancel(AvecP xx DvecR)-PvecR xx AvecP-PvecRxxDvecR-cancel(DvecRxxAvecP)+DvecRxxPvecR|`
`=(1/2)|-2PvecRxxAvecP-2PvecRxxDvecR|` => `S_(ABCD)=|PvecR xx (AvecP+DvecR)|`

Quadrilateral PQRS
`SvecQ=SvecP+PvecQ`

`PvecD=PvecS+SvecD` `->` note that `SvecD=2AvecS`
`PvecD=PvecS+2AvecS` `->`note that `AvecS=AvecP+PvecS`
`PvecD=PvecS+2(AvecP+PvecS)` => `3PvecS=(PvecD-2AvecP)` => `PvecS=(PvecD-2AvecP)/3` `->`note that `PvecD=PvecR+RvecD=PvecR-DvecR`
`-> PvecS=(PvecR-DvecR-2AvecP)/3`

`PvecC=PvecQ+QvecC` `->`note that `QvecC=(BvecQ)/2`
`PvecC=PvecQ+(BvecQ)/2` `->`note that `BvecQ=BvecP+PvecQ` (since `BvecP=-AvecP` => `BvecQ=-AvecP+PvecQ)`
`2PvecC=2PvecQ-AvecP+PvecQ` => `3PvecQ=2PvecC+AvecP` => `PvecQ=(2PvecC+AvecP)/3` `->`note that `PvecC=PvecR+RvecC=PvecR+DvecR`
`PvecQ=(2PvecR+2DvecR+AvecP)/3`

Back to `SvecQ`
`SvecQ=SvecP+PvecQ=-PvecS+PvecQ`
`SvecQ=(-PvecR+DvecR+2AvecP)/3+(2PvecR+2DvecR+AvecP)/3`
`SvecQ=(PvecR)/3+DvecR+AvecP`

`S_(PQRS)=(1/2)*|PvecR xx SvecQ|=(1/2)|PvecR xx ((PvecR)/3+DvecR+AvecP)|`
`S_(PQRS)=(1/2)*|PvecR xx (DvecR+AvecP)|`

Finally,
`S_(PQRS)/S_(ABCD)=((1/2)*cancel(|PvecR xx (DvecR+AvecP)|))/cancel(|PvecR xx (DvecR+AvecP)|)=1/2`
Q.E.D.