Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
MediumIn the figure,$ABCD$ is a quadrilateral and $BE \parallel AC$. $BE$ meets $DC$ produced at $E$. Show that the area of $\Delta ADE$ is equal to the area of the quadrilateral $ABCD$.
(N/A) Observe the figure carefully.
$\Delta BAC$ and $\Delta EAC$ lie on the same base $AC$ and between the same parallel lines $AC$ and $BE$.
Therefore,$ar(\Delta BAC) = ar(\Delta EAC)$ (Triangles on the same base and between the same parallels are equal in area).
Now,add $ar(\Delta ADC)$ to both sides of the equation:
$ar(\Delta BAC) + ar(\Delta ADC) = ar(\Delta EAC) + ar(\Delta ADC)$
From the figure,$ar(\Delta BAC) + ar(\Delta ADC) = ar(ABCD)$ and $ar(\Delta EAC) + ar(\Delta ADC) = ar(\Delta ADE)$.
Therefore,$ar(ABCD) = ar(\Delta ADE)$.
$\Delta BAC$ and $\Delta EAC$ lie on the same base $AC$ and between the same parallel lines $AC$ and $BE$.
Therefore,$ar(\Delta BAC) = ar(\Delta EAC)$ (Triangles on the same base and between the same parallels are equal in area).
Now,add $ar(\Delta ADC)$ to both sides of the equation:
$ar(\Delta BAC) + ar(\Delta ADC) = ar(\Delta EAC) + ar(\Delta ADC)$
From the figure,$ar(\Delta BAC) + ar(\Delta ADC) = ar(ABCD)$ and $ar(\Delta EAC) + ar(\Delta ADC) = ar(\Delta ADE)$.
Therefore,$ar(ABCD) = ar(\Delta ADE)$.
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View SolutionIn the figure,diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at $O$ such that $OB = OD$. If $AB = CD$,then show that:
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