Question

In the given figure P is a point in the interior of a llgm ABCD ,show that
I) $ar\left(APB\right)+ar\left(PCD\right)=\frac{1}{2}ar\left(ABCD\right)$
II) $ar\left(APD\right)+ar\left(PBC\right)=ar\left(APB\right)+ar\left(PCD\right)$
[hint :through P ,drawn a line parallel to AB]

Answer 1

Given: ABCD is a parallelogram. Let MN passing through P be parallel to AB.

Since, Area of triangle is half of the area of parallelogram if they lie on same base and between same parallel lines.

$\therefore ar\left(△APB\right)=\frac{1}{2}ar\left(||gmABNM\right)....\left(1\right)$

$\therefore ar\left(△DPC\right)=\frac{1}{2}ar\left(||gmDCNM\right)....\left(2\right)$

Adding equation 1 and 2, we have ,

$ar\left(△APB\right)+ar\left(△DPC\right)=\frac{1}{2}ar\left(||gmABNM\right)+\frac{1}{2}ar\left(||gmDCNM\right)$

$\Rightarrow ar\left(△APB\right)+ar\left(△DPC\right)=\frac{1}{2}ar\left(||gmABCD\right)$

Similarly, if we take XY passing through P to be parallel to AD.

Then,

$ar\left(△APD\right)=\frac{1}{2}ar\left(||gmAXYD\right)....\left(3\right)$

$ar\left(△BPC\right)=\frac{1}{2}ar\left(||gmBCYX\right)....\left(4\right)$

Adding equation 3 and 4, we have ,

$ar\left(△APD\right)+ar\left(△BPC\right)=\frac{1}{2}ar\left(||gmAXYD\right)+\frac{1}{2}ar\left(||gmBCYX\right)$

$\Rightarrow ar\left(△APD\right)+ar\left(△BPC\right)=\frac{1}{2}ar\left(||gmABCD\right)$