$A D$ and $B C$ are equal perpendiculars to a line segment $A B$ (See the given figure). Show that $C D$ bisects $A B$.
Solution:
In $\triangle \mathrm{BOC}$ and $\triangle \mathrm{AOD}$,
$\angle B O C=\angle A O D$ (Vertically opposite angles)
$\angle C B O=\angle D A O\left(\right.$ Each $\left.90^{\circ}\right)$
$\mathrm{BC}=\mathrm{AD}$ (Given)
$\therefore \triangle \mathrm{BOC} \cong \triangle \mathrm{AOD}(\mathrm{AAS}$ congruence rule $)$'
$\therefore \mathrm{BO}=\mathrm{AO}(\mathrm{By} \mathrm{CPCT})$
$\Rightarrow \mathrm{CD}$ bisects $\mathrm{AB}$.
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