# Q: 9         The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. $\small 9.26$). Show that  $\small ar(ABCD)=ar(PBQR)$.                 [Hint: Join AC and PQ. Now compare  $\small ar(ACQ)$ and  $\small ar(APQ)$.]

M manish

Join the AC and PQ.

It is given that ABCD is a ||gm and AC is a diagonal of ||gm
Therefore, ar($\Delta$ABC) = ar($\Delta$ADC) = 1/2 ar(||gm ABCD).............(i)

Also, ar($\Delta$PQR) = ar($\Delta$BPQ) = 1/2 ar(||gm PBQR).............(ii)

Since $\Delta$AQC and $\Delta$APQ are on the same base AQ and between same parallels AQ and CP.
$\therefore$  ar($\Delta$AQC) = ar ($\Delta$APQ)

Now, subtracting $\Delta$ABQ from both sides we get,

ar($\Delta$AQC) - ar ($\Delta$ABQ) = ar ($\Delta$APQ) - ar ($\Delta$ABQ)
ar($\Delta$ABC) = ar ($\Delta$BPQ)............(iii)

From eq(i), (ii) and (iii) we get

$\small ar(ABCD)=ar(PBQR)$

Hence proved.

Exams
Articles
Questions