A radius drawn to bisect a chord in a circle will always meet the chord at 90 degrees.

Use a mathematical method (that does not use vectors), to prove this.

I've never been too great with geometric proofs (Headbang)

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- October 4th 2008, 06:49 AMDJ HoboGeometry proof
**A radius drawn to bisect a chord in a circle will always meet the chord at 90 degrees.**

Use a mathematical method (that does not use vectors), to prove this.

I've never been too great with geometric proofs (Headbang) - October 4th 2008, 07:14 AMskeeter
let the chord be segment AB. let the circle center be point O.

let M be the intersection point of the bisecting radius and chord AB.

M is the midpoint of AB ... why?

now, prove that triangle OMA is congruent to triangle OMB, then use corresponding parts of congruent triangles to show that angle OMA is congruent to angle OMB.

from this point, it should be easy to show that OM is perpendicular to AB. - October 4th 2008, 07:22 AMDJ Hobo
- October 4th 2008, 07:37 AMskeeter
formally, you can't just say M is the midpoint of AB ... the reason must be given, i.e. "definition of a segment bisector".

- October 4th 2008, 07:39 AMDJ Hobo
- October 4th 2008, 08:10 AMskeeter
you do understand that every statement in a proof requires a reason, correct?

that's all I'm trying to tell you. - October 4th 2008, 08:12 AMDJ Hobo