ABCD is a trapezium with AB parallel to DC. A line parallel to AC… - Vidyadip

Question

$\text{ABCD}$ is a trapezium with $AB$ parallel to $DC. A$ line parallel to $AC$ intersects $AB$ at $X$ and $BC$ at $Y.$
Prove that the area of $\triangle ADX =$ area of $\triangle ACY.$

✓

Answer

Join $CX, DX$ and$ AY.$

Now, triangles $ ADX$ and $ACX$ are on the same base $AX$ and between the parallels $AB$ and $DC.$
$\therefore A( \triangle ADX ) = A( \triangle ACX ) ….(i)$
Also, triangles $ACX$ and $ACY$ are on the same base $AC$ and between the parallels $AC$ and $XY.$
$\therefore A( \triangle ACX ) = A( \triangle ACY ) ….(ii)$
From $(i)$ and $(ii),$ we get
$A( \triangle ADX ) = A( \triangle ACY )$

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