3 Tutor System
Starting just at 265/hour

# 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 Figure). Show that ar (ABCD) = ar (PBQR).[Hint : Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]

Construction: Let us join AC and PQ.

$$\triangle{ACQ}$$ and $$\triangle{AQP}$$ are the same base AQ and between the same parallels AQ and CP.

$$\therefore$$ Area $$\triangle{ACQ}$$ = Area $$\triangle{AQP}$$
$$\Rightarrow Area (\triangle{ACQ} ) - Area (\triangle{ABQ} ) = Area (\triangle{APQ} ) - Area (\triangle{ABQ})$$
$$\Rightarrow$$ Area $$(\triangle{ABC}$$) = Area ($$\triangle{QBP}$$) ...(i)

$$\because$$ AC and PQ are diagonals of parallelograms ABCD and PBQR, respectively
$$\therefore$$ Area ($$\triangle{ABC}$$) = ($$\frac{1}{2}$$ ) Area (ABCD) ...(ii)

Also, Area ($$\triangle{QBP}$$) = ($$\frac{1}{2}$$ ) Area (PBQR) ...(iii)

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

($$\frac{1}{2}$$ ) Area (ABCD) = ($$\frac{1}{2}$$ ) Area(PBQR)
$$\therefore$$ Area (ABCD) = Area(PBQR)
Hence, proved.