Question

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♡$\underline{ \underline{ \text{Question : }}}$ In the given figure , L and M are the mid-points of two equal chords AB and CD of a circle with centre O. Prove that :
i.
$\angle$OLM = $\angle$ OML

ii. $\angle$ ALM = $\angle$ CML

Answer 1

Answer:

given chord AB=chord CD

L and M are the mid-points of two equal chords AB and CD of a circle

we have

(Equal chords are equidistant from the centre)

In ∆ OLM

OL = OM

<OLM= <OMLbase angles

opposite to equal sides of a A

But <OLA = <OMC(Each = 90°being perpendicular )

On adding

<OLM+<OLA = <OML+<OMC

=<ALM = <CML

Answer 2

Answer:

Given: chord AB=chord CD

L and M are the mid-points of two equal chords AB and CD of a circle

we have

(Equal chords are equidistant from the centre)

In ∆ OLM

OL = OM

<OLM= <OMLbase angles

opposite to equal sides of a A

<OLA = <OMC(Each = 90°being perpendicular) Adding

<OLM+<OLA = <OML+<OMC

<ALM=<CML

Hence proved.