Straight Lines which make Equal Angles with Perpendicular to Straight Line are Equal

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Theorem

Let $AB$ be a straight line.

Let $C$ be a point which is not on $AB$.

Let $D$ be a point on $AB$ such that $CD$ is perpendicular to $AB$.

Let $E, F$ be points on $AB$ such that $\angle DCE = \angle DCF$.


Then $CE = CF$.


Proof

$\triangle CDE$ and $\triangle CDF$ are right triangle where $CE$ and $CF$ are the hypotenuses.

We have:

$\angle CDE = \angle CDF$ as both are right angles.
$\angle DCE = \angle DCF$ by hypothesis.
$CD$ is common.

Thus by Triangle Angle-Side-Angle Congruence, $\triangle CDE$ and $\triangle CDF$ are congruent.

Hence the result.

$\blacksquare$


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