Question

In the given figure, P is any point on the point on the diagonal AC of the parallelogram ABCD. Show that $ar\left( {\Delta ADP} \right) = ar\left( {\Delta ABP} \right)$

Answer 1

Hint:
It is given in the question that P is any point on the point on the diagonal AC of the parallelogram ABCD.
So, we will show that $ar\left( {\Delta ADP} \right) = ar\left( {\Delta ABP} \right)$
we know that the diagonals of a parallelogram bisect each other.

Complete step by step solution:
It is given in the question that P is any point on the point on the diagonal AC of the parallelogram ABCD.
So, we have to show that $ar\left( {\Delta ADP} \right) = ar\left( {\Delta ABP} \right)$
Let, the diagonal AC and DB intersect at O.
Since, we know that the diagonals of a parallelogram bisect each other
So, O is the midpoint of both AC and DB.
We know that the median of a triangle bisects each other.
In $\Delta DBP,$
PO is median of $\Delta DBP$
 $\therefore ar\left( {\Delta DPO} \right) = ar\left( {\Delta POB} \right)$ (I)
Now, In $\Delta ADB,$
AO is median of $\Delta ADB$
 $\therefore ar\left( {\Delta ADO} \right) = ar\left( {\Delta AOB} \right)$ (II)
Now, subtract equation (II) from equation (I)
$\therefore ar\left( {\Delta ADO} \right) - ar\left( {\Delta DPO} \right)$ $ = $ $ar\left( {\Delta AOB} \right) - ar\left( {\Delta POB} \right)$
$\therefore ar\left( {\Delta ADP} \right) = ar\left( {\Delta ABP} \right)$
Hence proved

Note:
Parallelogram: A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measures. The 3-dimensional counterpart of a parallelogram is a parallelepiped.