Mathematics
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
Answer
(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.
Related Questions
Prove that diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.
Prove that a diameter of a circle, which bisects a chord of the circle, also bisects the angle subtended by the chord at the centre of the circle.
In the given figure, AB and AC are equal chords of a circle with centre O and OP ⟂ AB, OQ ⟂ AC. Prove that PB = QC.
In an equilateral triangle, prove that the centroid and the circumcentre of the triangle coincide.