# prove perpendicular bisector of the chord of a circle goes though the origin

Hi

Can you show me how to prove that the perpendicular bisector of a chord goes through the origin, or from another perspective, that if a line bisects a chord and goes through the origin it is perpendicular to the chord. I have looked for examples on the internet that describe how to do this but they all use something called the perpendicular bisector theorem. I have not covered this yet in my book. This is just the beginning of a basic geometry section.

I saw this example here

and i understood it up to this line m (angle) CMA

I don't know what the notation with the little m means. I also don't know the rule used in this angle that 2 angles that are supplementary and complementary must both be 90 degrees.

I guess the problem is explaining it using proofs that I have covered and y ou don't know what I do and don't know yet!

Many thanks
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Commented:
>> that if a line bisects a chord and goes through the origin it is perpendicular to the chord.

think triangles. Draw a circle on a piece of paper, and draw a chord on it. Then draw a line from the center of that chord to the center of the circle, and draw two radii that go towards the extremes of the chord.

You have now drawn two triangles. Since the two radii have the same length (it's a circle), the common triangle side is the same (since it's common), and the remaining two triangle sides have the same length (since it's a bisector), we know that both triangles have the same form, and thus also have the same (corresponding) angles.

So, the two angles made by the bisector with the chord must be the same. The only way that that is possible, is for both angles to be 90 degrees (since the sum of both angles is 180 degrees).
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Commented:
>>  that 2 angles that are supplementary and complementary must both be 90 degrees.

supplementary and congruent (not complementary).
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Author Commented:
Many thanks
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