Question

P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP,
Show that $ar\left(\Delta PBQ\right)=ar\left(\Delta ARC\right)$. [3 MARKS]

Answer 1

Concept : 0.5 mark
Application : 0.5 mark
Proof : 2 marks

Take a point S on AC such that S is the mid-point of AC. Extend PQ to T such that PQ = QT.
Join TC, QS, PS and AQ.

In $\Delta ABC$, P and Q are the mid-points of AB and BC respectively.
$PQ||ACandPQ=\frac{1}{2}AC$ [mid-point theorem]
$\therefore PQ||ASandPQ=AS$ (As S is the mid-point of AC)
$\Rightarrow PQSA$ is a parallelogram.

We know that diagonals of a parallelogram bisect it into equal areas of triangles.
$\therefore ar\left(\Delta PAS\right)=ar\left(\Delta SQP\right)=ar\left(\Delta PAQ\right)=ar\left(\Delta SQA\right)$

Similarly,
$ar\left(\Delta PSQ\right)=ar\left(\Delta CQS\right)$ (For parallelogram PSCQ)
$ar\left(\Delta QSC\right)=ar\left(\Delta CTQ\right)$ (For parallelogram QSCT)
$ar\left(\Delta PSQ\right)=ar\left(\Delta QBP\right)$ (For parallelogram PSQB)

Thus,
$ar\left(\Delta ABC\right)=4ar\left(\Delta PBQ\right)....\left(1\right)$

$\Rightarrow ar\left(\Delta PBQ\right)=\frac{1}{4}ar\left(\Delta ABC\right)...\left(2\right)$

In parallelogram PACT,
$ar\left(\Delta ARC\right)=\frac{1}{2}ar\left(\Delta PAC\right)\left(CRisthemedianof\Delta PAC\right)$
$=\frac{1}{2}\times \frac{1}{2}ar\left(PCisthediagonalofparallelogramPACT\right)$
$=\frac{1}{4}ar\left(\Delta PACT\right)$
$=\frac{1}{4}ar\left(\Delta ABC\right)$
$=ar\left(\Delta PBQ\right)$