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.
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