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# $ \mathrm{AD} $ and $ \mathrm{BC} $ are equal perpendiculars to a line segment $ \mathrm{AB} $ (see Fig. 7.18). Show that $ \mathrm{CD} $ bisects $ \mathrm{AB} $.

"

**Given:**

$AD$ and $BC$ are equal perpendiculars to a line segment $AB$.

**To do:**

We have to show that $CD$ bisects $AB$.

**Solution:**

We know that,

From Angle Angle Side congruence rule:

If two pairs of corresponding angles along with the opposite or non-included sides are equal to each other then the two triangles are said to be congruent.

Therefore,

$\triangle AOD \cong \triangle BOD$

$\angle A=\angle B$ (since they are perpendicular to each other)

Given,

$AD=BC$

We know that,

vertically opposite angles are equal,

This implies,

$\angle AOD=\angle BOD$

Therefore,

$\triangle AOD \cong \triangle BOC$

We also know that,

From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding angles and sides must be equal.

This implies,

$AO=OB$

Therefore,

$CD$ bisects $AB$.

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