Question

In Figure, P is a point in the interior of a parallelogram ABCD. Show that
$\text{(i)}ar\left(APB\right)+ar\left(PCD\right)=\frac{1}{2}ar\left(ABCD\right)$ $\text{(ii)}ar\left(APD\right)+ar\left(PBC\right)=ar\left(APB\right)+ar\left(PCD\right)$ [Hint : Through P, draw a line parallel to AB.]

Answer 1

(i) A line GH is drawn parallel to AB passing through P.
In a parallelogram,
AB || GH (by construction) — (i)
$\therefore AD||BC\Rightarrow AG||BH—\left(ii\right)$
From equations (i) and (ii),
ABHG is a parallelogram.
Now,
$△APB\text{and parallelogram ABHG are lying on the same base AB and in-between the same parallel lines AB and GH.}$
$\therefore ar\left(△APB\right)=\frac{1}{2}ar\left(ABHG\right)—\left(iii\right)$
also,
$△PCD\text{and parallelogram CDGH are lying on the same base CD and in-between the same parallel lines CD and GH.}$
$\therefore ar\left(△PCD\right)=\frac{1}{2}ar\left(CDGH\right)—\left(iv\right)$
Adding equations (iii) and (iv),
$ar\left(△APB\right)+ar\left(△PCD\right)=\frac{1}{2}\left[ar\right(ABHG)+ar(CDGH\left)\right]$
$\Rightarrow ar\left(APB\right)+ar\left(PCD\right)=\frac{1}{2}ar\left(ABCD\right)$ (ii) A line EF is drawn parallel to AD passing through P.
In the parallelogram,
AD || EF (by construction) — (i)
$\therefore AB||CD\Rightarrow AE||DF—\left(ii\right)$
From equations (i) and (ii),
AEDF is a parallelogram.
Now,
$△APD\text{and parallelogram AEFD are lying on the same base AD and in-between the same parallel lines AD and EF.}$
$\therefore ar\left(△APD\right)=\frac{1}{2}ar\left(AEFD\right)—\left(iii\right)$
also,
$△PBC\text{and parallelogram BCFE are lying on the same base BC and in-between the same parallel lines BC and EF.}$
$\therefore ar\left(△PBC\right)=\frac{1}{2}ar\left(BCFE\right)—\left(iv\right)$
Adding equations (iii) and (iv),
$ar\left(△APD\right)+ar\left(△PBC\right)=\frac{1}{2}ar\left(AEFD\right)+ar\left(BCFE\right)$
$\Rightarrow ar\left(APD\right)+ar\left(PBC\right)=ar\left(APB\right)+ar\left(PCD\right)$