Question
$ABCD$ is a parallelogram whose diagonals $AC$ and $BD$ intersect at $O$. A line through $O$ intersects $AB$ at $P$ and $DC$ at $Q$. Prove that $\text{ar}(\triangle\text{POA})=\text{ar}(\triangle\text{QOC}).$
✓
Answer
In triangles $POA$ and $QOC$, we have $\angle\text{AOP}=\angle\text{COQ}$
$\text{AO}=\text{OC}$
$\angle\text{PAC}=\angle\text{QCA}$
So, by $ASA$ congruence criterion,
we have $\triangle\text{POA}\cong\triangle\text{QOC}$
$\Rightarrow\text{ar}(\triangle\text{POA})=\text{ar}(\triangle\text{QOC}).$
$\text{AO}=\text{OC}$
$\angle\text{PAC}=\angle\text{QCA}$
So, by $ASA$ congruence criterion,
we have $\triangle\text{POA}\cong\triangle\text{QOC}$
$\Rightarrow\text{ar}(\triangle\text{POA})=\text{ar}(\triangle\text{QOC}).$
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