If the angle - bisector of two intersecting chords of a circle passes through its centre , then prove that the chords are equal .

Explanation

Let the two chords AB and AC of the circle with centre at O intersect each other at A. The bisector AO of the internal angle `angle BAC` passes through the centreO.
We have to prove that AB = AC .
Construction : let us join O, B and O,C .
Proof : In `Delta AOB` and `Delta AOC`
`angle OBA = angle OAB [ because OB = OA ` = radii of same circle ]
` = angle OCA [ because angle OAB = angle OAC ` given ]
` = angle OCA [ because OC = OA = " radii of same circle"]`
`i.e., angle OBA = angle OCA and angle OAB = angle OAC`
and OA is common to both .
By the condition of A - A - S congreucency of congruent triagle .
` Delta AOB ~= Delta AOC`
`therefore AB = AC [ because ` similar sides of congruency triangles].
`therefore` the chord AB and AC are equal .