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?

1 Answer
Feb 8, 2016

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

Explanation:

I created this figure using MS ExcelI created this figure using MS Excel

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.