$ \mathrm{AD} $ and $ \mathrm{BC} $ are equal perpendiculars to a line segment $ \mathrm{AB} $ (See the given figure). Show that CD bisects AB.
"
Given:
$AD=BC$
$AD \perp AB$
$BC \perp AB$
To do:
We have to show that CD bisects AD.
Solution:
In $\triangle ADO$ and $\triangle BCO$,
$\angle OAD=\angle OBC=90^o$
$AD=BC$ (Given)
$\angle DOA=\angle COB$ (Vertically opposite angles)
Therefore,
$\triangle ADO \cong\ \triangle BCO$ (By AAS congruence)
This implies,
$OB=OA$ (CPCT)
Hence proved.
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