Question

$ABCD$ is a trapezium with $AB\parallel DC$. A line parallel to $AC$ intersects $AB$ at $X$ and $BC$ at $Y$. Prove that $ar\left(△ADX\right)=ar\left(△ACY\right)$

Answer 1

Given : $ABCD$ is a trapezium with $AB\parallel DC$ and $XY\parallel AC.$

Join $YA,XC$ and $XD.$

Triangles $ACX$ and $ACY$ have same base $AC$ and are between same parallels $AC$ and $XY$

So, $ar\left(△ACX\right)=ar\left(△ACY\right)$ ......(i)

Triangles $ACX$ and $ADX$ have same base $AX$ and are between same parallels $AB$ and $DC.$

So, $ar\left(△ACX\right)=ar\left(△ADX\right)$ ......(ii)

From (i) and (ii),

$ar\left(△ADX\right)=ar\left(△ACY\right)$