Question

Diagonals AC and BD of a trapezium ABCD with $AB\parallel DC$ intersect each other at O. Prove that $ar(AOD) = ar(BOC)$ .

Answer 1

Hint: We will make use of the relation between areas of triangles which have the same base and lie between the same parallel lines. That is, the area of two triangles with the same base and lying between the same parallel lines is always equal.

Complete Step-by-step Solution
Given: ABCD is a trapezium with $AB\parallel DC$ and diagonals AC and BD intersect each other at O.
Now, triangle ADC and triangle BDC lie on the same base DC and between the same parallels AB and CD.
$ar(\Delta ADC) = ar(\Delta BDC)$
Subtract area of triangle DOC from both sides.
$\begin{array}{l}
ar(\Delta ADC) - ar(\Delta DOC) = ar(\Delta BDC) - ar(\Delta DOC)\\
ar(\Delta AOD) = ar(\Delta BOC)
\end{array}$
Therefore, the required relation $ar(\Delta AOD) = ar(\Delta BOC)$ is proved.
Note:In this problem, the theorem of triangles is used which states that when two triangles lie on the same base and between same parallel lines then both the triangles are equal in area. To prove $ar(\Delta ADC) = ar(\Delta BDC)$, we will use this theorem. Here, you could also take the area of triangles ADB and ACB equal and then subtract the area of triangle AOB.