Question

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

[Hint : Join $\mathrm{CX}$.]

Answer 1

proving that $\mathrm{ar}(△\mathrm{ADX})=\mathrm{ar}(△\mathrm{ACY})$:
Given:

$\mathrm{ABCD}$ is a trapezium with,$\mathrm{AB}\parallel \mathrm{DC},\mathrm{XY}\parallel \mathrm{AC}$

Construction:

Join $\mathrm{CX}$

Proof:

$\mathrm{ar}\left(\mathrm{\Delta ADX}\right)=\mathrm{ar}\left(\mathrm{\Delta AXC}\right)$————($1$) (They are on the same base$\mathrm{AX}$ and in-between the same parallels $\mathrm{AB}$ and $\mathrm{CD}$)

$\mathrm{ar}\left(\mathrm{\Delta AXC}\right)=\mathrm{ar}\left(\mathrm{\Delta ACY}\right)$_______________($2$) (They are on the same base $\mathrm{AC}$ and in-between the same parallels $\mathrm{XY}$ and $\mathrm{AC}$)

From equations ($1$) and ($2$),

$\mathrm{ar}\left(\mathrm{\Delta ADX}\right)=\mathrm{ar}\left(\mathrm{\Delta ACY}\right)$

Hence we prove that $\mathrm{ar}\left(\mathrm{\Delta ADX}\right)=\mathrm{ar}\left(\mathrm{\Delta ACY}\right)$