Question

E is the mid-point of the non-parallel side BC of a trapezium ABCD. E is joined to the opposite vertices A and D prove that $\Delta ABE+\Delta DCE=\frac{1}{2}$ trapezium ABCD.

Answer 1

Given
ABCD is a trapezium such that $AB||DC$.
E is the mid-point of BC
To prove: Area $\left(\Delta ABE\right)+area\left(\Delta DEC\right)=\frac{1}{2}\times area\left(trapeziumABCD\right)$
Proof:
$Area\left(\Delta ABE\right)=\frac{1}{2}\times area\left(\Delta ABC\right)$ ....(1)
[since the median divides the triangle into equal parts]
Similarly, $area\left(\Delta DEC\right)=\frac{1}{2}\times area\left(\Delta BDC\right)$ ....(2)
$Area\left(\Delta BDC\right)=area\left(\Delta ADC\right)$
[triangles formed between same pair of parallel lines and with the same base are equal in areas]
Therefore,
$Area\left(\Delta DEC\right)=\frac{1}{2}\times area\left(\Delta ADC\right)$ ....(3)
Adding eq(1) and eq(3)
$Area\left(\Delta ABE\right)+area\left(\Delta DEC\right)=\frac{1}{2}\times \left[area\right(\Delta ABC)+area(\Delta ADC\left)\right]=\frac{1}{2}\times areatrapeziumABCD$