Question

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

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Solution

## proving that $\mathrm{ar}\left(△\mathrm{ADX}\right)=\mathrm{ar}\left(△\mathrm{ACY}\right)$: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 $\mathrm{ar}\left(\mathrm{\Delta ADX}\right)=\mathrm{ar}\left(\mathrm{\Delta ACY}\right)$

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