If two equal chords intersect within the circle,prove that the line joining the point of intersection to the centre makes equal angles with the chords.

given: AB and CD are the two equal chords of a circle with centre O.

let AB and CD intersect at point M.

TPT: the line joining the point of intersection of chords to the center makes equal angles with the chords.

i.e. ∠OMA=∠OMD

construction: draw OP⊥AB and OQ⊥CD

proof:

in the triangle OPM and triangle OQM,

OP=OQ [equal chords are equally distant from the center]

∠OPM=∠OQM [by the construction]

OM is common .

therefore by right angle hypotenuse side congruency triangles are congruent.

therefore by the property of congruency,

∠OMP=∠OMQ

i.e. ∠OMA=∠OMD

thus the line joining the point of intersection of chords to the center makes equal angles with the chords.

hope this helps you.

cheers!!

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