$AD$ and $BC$ are equal perpendiculars to a line segment $AB$ (see figure). Show that $CD$ bisects $AB$.
(N/A) Given: $AD \perp AB$,$BC \perp AB$ and $AD = BC$.
To prove: $CD$ bisects $AB$,i.e.,$OA = OB$.
Proof:
In $\Delta OBC$ and $\Delta OAD$:
$1$. $\angle OBC = \angle OAD = 90^\circ$ (Given)
$2$. $\angle BOC = \angle AOD$ (Vertically opposite angles)
$3$. $BC = AD$ (Given)
By $AAS$ congruence rule,$\Delta OBC \cong \Delta OAD$.
Therefore,$OB = OA$ (by $c.p.c.t.$).
Since $OB = OA$,$O$ is the midpoint of $AB$.
Hence,$CD$ bisects $AB$.
To prove: $CD$ bisects $AB$,i.e.,$OA = OB$.
Proof:
In $\Delta OBC$ and $\Delta OAD$:
$1$. $\angle OBC = \angle OAD = 90^\circ$ (Given)
$2$. $\angle BOC = \angle AOD$ (Vertically opposite angles)
$3$. $BC = AD$ (Given)
By $AAS$ congruence rule,$\Delta OBC \cong \Delta OAD$.
Therefore,$OB = OA$ (by $c.p.c.t.$).
Since $OB = OA$,$O$ is the midpoint of $AB$.
Hence,$CD$ bisects $AB$.
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