$A D$ 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 $\mathrm{AB}$.