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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.