Hello, can someone please give me a simple proof to the following theorem:
"The perpendicular bisector a chord passes through the centre of the circle."
I have attached a diagram of what I mean and web link of a proof that I did not understand below.
https://proofwiki.org/wiki/Perpendicular_Bisector_of_Chord_Passes_Through_Center
Please explain simply and fully because I have an exam on this tomorrow. Also, could you explain the converse theorem whereby a bisector passes through the centre of the circle, prove it's perpendicular and a perpendicular line passes through the centre, prove it bisects the chord.
Best Answer
Actually, instead of using SAS criterion, you can use SSS criterion. $$AC = BC$$ [∵ AC and BC are radii] $$CD = BD$$ [∵ BC is bisected to form CD and BD] $$DC = DC$$ [Common]
If the perpendicular bisector does not pass through C, then the hypotenuse sides would not be radii, therefore the triangles won't be congruent, therefore the bisector won't be perpendicular, which is a proof by contradiction.
For more information visit https://www.khanacademy.org/math/geometry-home/cc-geometry-circles/area-inscribed-triangle/v/sss-to-show-a-radius-is-perpendicular-to-a-chord-that-it-bisects and https://www.khanacademy.org/math/geometry-home/cc-geometry-circles/area-inscribed-triangle/v/perpendicular-radius-bisects-chord