Question
$\text{AD}$ and $\text{BC}$ are equal perpendiculars to a line segment $\text{AB}$. If $\text{AD}$ and $\text{BC}$ are on different sides of $\text{AB}$ prove that $\text{CD}$ bisects $\text{AB}.$

Answer



In $\triangle AOD$ and $\triangle BOC$,
$\angle \mathrm{AOD}=\angle \mathrm{BOC} \quad \ldots. ($vertically opposite angles$)$
$\angle \mathrm{DAO}=\angle \mathrm{CBO} \quad \ldots . .\left(\right.$ each $\left.90^{\circ}\right)$
$\text{AD}=\text{BC} \quad \quad \ldots. ($given$)$
$\therefore \triangle \mathrm{AOD} \cong \triangle \mathrm{BOC} \quad...($by $\text{AAS}$ congruence criterion$)$
$\Rightarrow \text{AO}=\text{BO} ...($c.p.c.t$.)$
$\Rightarrow O$ is the mid$-$point of $\text{AB}$. Hence, $\text{CD}$ bisects $\text{AB}$.

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