Hello, I need some help again. I need an algebraic proof of this theorem:
The perpendicular bisector of a chord of a circle passes through the center of the circle.
There are plenty of geometric proofs out there, but I need one that just makes use of the midpoint formula and the Pythagorean theorem. I am just having some conceptual difficulty because I do not know how you prove that a point is on a certain line. I think I should show that the midpoint of the perpendicular bisector equals the midpoint of the diameter, but I don't know how to get a diameter. If someone could outline the basic steps you might use, I would really appreciate it!
Draw a circle and a chord.Erect the perpendicular bisector of chord AB is chord CD is the PBisector E is midpoint of AB. Connect all points. You have a quadrangle ACBD.
Triangle ACE is congruent to CEB. ASA CE is common Angle CEA is 90 d and AE=BE. Triangles AED and BED are congruent in like manner.Show that angles CAD = CBD.Opposite angles of an insribed quad are supplementry making CAD and CBD = 90 degrees. CD is a diameter so passes thru center
Plato, thank you so much for your help. When I put the numbers into the formula y = mx + b and set x=0, I get
b(b-d)
-------
a(a-c)
whereas I was hoping it would cancel to zero and that would prove it. Is this what you meant? Perhaps I am making a mistake.