Question

$ABCD$ is a trapezium with $AB\parallel DC$. A line parallel to $BD$ intersects $CD$ at $X$ and $BC$ at $Y$ such that $BD\parallel XY$. Prove that $ar\left(ADX\right)=ar\left(BDY\right)$.

Answer 1

Given ABCD is a trapezium where $AB\parallel DC.$

$BD$ intersect $CD$ at $X$ and $BC$ at $Y$ i.e, $BD\parallel XY.$

We have to prove that $ar\left(ADX\right)=ar\left(BDY\right)$

For $△BDX$ and $△BDY$ lie on the same base $BD$ and are between parallel lines $BD$ and $XY.$

$\therefore ar\left(△BDX\right)=ar\left(△BDY\right)$

[ triangle with same base and between the same parallels are equal in area. ]

For $△BDX$ and $△ADX$ lie on the same base $DX$ and are between parallel lies $AB$ and $DX.$

[ As $AB\parallel DC,$ parts of parallel lines are parallel ]

$\therefore ar\left(△BDX\right)=ar\left(△ADX\right)$

Triangles with same base and between the same parallel are equal in area.

$\Rightarrow ar\left(△ADX\right)=ar\left(△BDY\right)$

Hence, the answer is proved.