In the given figure, L and M are mid-points of two equal chords AB and CD of a circle with centre O. Prove that :

(i) ∠OLM = ∠OML

(ii) ∠ALM = ∠CML

(i) Given,

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

We know that,

A straight line drawn from the center of a circle to bisect a chord, which is not a diameter, is at right angles to the chord.

∴ OL ⊥ AB and OM ⊥ CD.

Since, AB and CD are equal chords and perpendicular drawn from center to equal chords are equal in length.

∴ OL = OM

Thus, in triangle OLM,

∴ ∠OLM = ∠OML (Angles opposite to equal sides are equal)

Hence, proved that ∠OLM = ∠OML.

(ii) We have,

∠OLM = ∠OML

∠OLA - ∠ALM = ∠OMC - ∠CML

Since, ∠OLA and ∠OMC both equal to 90°.

∴ ∠ALM = ∠CML.

Hence, proved that ∠ALM = ∠CML.