Question
✓
Answer
$ABCD$ is a parallelogram.
$\therefore AD \| BC$ and $AB \| DC$
Consider $\triangle ABC$ and $\triangle BDC$.
Both the triangles are drawn in two parallel lines. Hence the distance between the two parallel lines is the height of both triangles.
In $\triangle ABC$ and $\triangle BDC$, common base is $BC$ and heights are equal.
Hence, $A (\triangle ABC )= A (\triangle BDC )$
In $\triangle A B C$ and $\triangle A B D, A B$ is common base and and heights are equal.
$\therefore A (\triangle ABC )= A (\triangle ABD )$
$\therefore AD \| BC$ and $AB \| DC$
Consider $\triangle ABC$ and $\triangle BDC$.
Both the triangles are drawn in two parallel lines. Hence the distance between the two parallel lines is the height of both triangles.
In $\triangle ABC$ and $\triangle BDC$, common base is $BC$ and heights are equal.
Hence, $A (\triangle ABC )= A (\triangle BDC )$
In $\triangle A B C$ and $\triangle A B D, A B$ is common base and and heights are equal.
$\therefore A (\triangle ABC )= A (\triangle ABD )$
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