Question
Show that:A diagonal divides a parallelogram into two triangles of equal area.
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Answer
Suppose $ABCD$ is a parallelogram $...($given$)$
Consider the triangles $ABC$ and $ADC :$
$AB = CD ......[ \text{ABCD}$ is a parallelogram$ ]$
$ADE = BC ......[\text{ABCD}$ is a parallelogram$ ]$
$AD = AD .....[$ common $]$
By Side$-$ Side $-$Side criterion of congruence, we have,
$\triangle ABC ≅ \triangle ADC$
Area of congruent triangles are equal.
Therefore, Area of $ABC =$ Area of $ADC$
Consider the triangles $ABC$ and $ADC :$
$AB = CD ......[ \text{ABCD}$ is a parallelogram$ ]$
$ADE = BC ......[\text{ABCD}$ is a parallelogram$ ]$
$AD = AD .....[$ common $]$
By Side$-$ Side $-$Side criterion of congruence, we have,
$\triangle ABC ≅ \triangle ADC$
Area of congruent triangles are equal.
Therefore, Area of $ABC =$ Area of $ADC$
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