Question

AB and CD are equal chords of a circle whose centre is O. OM is perpendicular to AB and ON is perpendicular to CD. Prove that angle OMN is equal to angle ONM

Answer 1

Answer:

∠OMN=∠ONM    (Hence Proved)

Step-by-step explanation:

Given: AB and CD are equal chords of a circle whose center is O. OM is perpendicular to AB and ON is perpendicular to CD.

To prove: ∠OMN=∠ONM

Figure: Please see the attachment

Theorem: If chords are equal of same circle then they are equidistance from center.

AB = CD

OM is distance of chord AB from center.

ON is distance of chord CD from center.

OM=ON (If chords are equal of same circle then they are equidistance from center)

In ΔMON, OM=ON

∠OMN=∠ONM  , In a triangle if opposite sides are equal then their corresponding angles are equal.

Hence proved

Answer 2

"To Prove:

∠OMN = ∠ONM

Solution:

Equidistant chord theorem states that “Chords equidistant from the centre of a circle are equal in length and vice versa.Here, AB = CD (equal chords)

From the diagram, OM is distance of chord AB from centre.

From the diagram, ON is distance of chord CD from centre.

OM = ON (If chords are equal of same circle then they are equidistance from centre)

In, ΔMON, OM = ON

∠OMN = ∠ONM, in a triangle if "opposite sides" are equal then their "corresponding angles" are "equal"."