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# If two chords of a circle are equally inclined to the diameter through their point of intersection, prove the chords are equal.

**Given:**

Two chords of a circle are equally inclined to the diameter through their point of intersection.

**To do:**

We have to prove that the chords are equal.**Solution:**

$AB$ and $AC$ are two chords which are equally inclined to the diameter $AD$ of the circle with centre $O$.

Draw $OL \perp AB$ and $OM \perp AC$.

In $\triangle OLA$ and $\triangle OMA$,

$\angle OLA=\angle OMA=90^o$

$OA=OA$ (Common side)

$\angle LAO=\angle MAO$ (Given)

Therefore,

\( \triangle \mathrm{OLA} \cong \triangle \mathrm{OMA} \) (By AAS congruence)

\( \Rightarrow \mathrm{OL}=\mathrm{OM} \) (CPCT)

This implies, the chords are equidstant from the centre $O$.

\( \Rightarrow A B=A C \) (Equal chords of a circle are equidistant from the centre of the circle)Hence proved.