Question

ABCD is trapezium with AB // DC. A line parallel to AC intersects AB at point M and BC at point N. then prove area of $\Delta ADM$ = area of $\Delta ACN$.

Answer 1

In the figure $X=M$ and $Y=N$

For $△ACM$ and $△ACN$ lie on the same base $AC$ and are between same parallel lines $AC$ and $MN$

$Ar\left(△ACM\right)=Ar\left(△ACN\right)\cdots \left(1\right)$

Also for $△ACM$ and $△ADM$ lie on the same base $AM$ and between parallel lines $AM$ and $DC$ [As $AB\parallel BC,$ parts of parallel lines are parallel $]$

$Ar(△ACM=Ar(△ADM)\cdots (2)$

From $\left(1\right)$ and $\left(2\right)$

$Ar\left(△ADM\right)=Ar\left(△ACN\right)$