Prove the perpendicular bisector of a chord passes through center

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!

Re: Prove the perpendicular bisector of a chord passes through center

Quote:

Originally Posted by

**Ragnarok** Hello, I need some help again. I need an *algebraic* proof of this theorem:

[B]The perpendicular bisector of a chord of a circle passes through the center of the circle.

Suppose the circle is $\displaystyle x^2+y^2=r^2$ (We can translate any circle to be centered at $\displaystyle (0,0).$

Say that $\displaystyle (a,b)~\&~(c,d)$ are points on the circle.

Then $\displaystyle \left( {\frac{{a + c}}{2},\frac{{b + d}}{2}} \right)$ is the midpoint of the cord.

The slope of the perpendicular bisector is $\displaystyle \frac{{a - c}}{{d - b}}$

Show that $\displaystyle (0,0)$ is on the perpendicular bisector.

Re: Prove the perpendicular bisector of a chord passes through center

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

Re: Prove the perpendicular bisector of a chord passes through 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.

Re: Prove the perpendicular bisector of a chord passes through center

Quote:

Originally Posted by

**Ragnarok** 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.

The equation of the bisector is $\displaystyle y-\frac{b+d}{2}=\frac{a-c}{d-b}\left(x-\frac{a+c}{2}\right)$.

Show that $\displaystyle (0,0)$ satisfies that equation,

Re: Prove the perpendicular bisector of a chord passes through center

Okay, putting 0 in for x and y I get:

$\displaystyle 0 - \frac{b+d}{2} = \frac{a-c}{d-b} \left (0 - \frac{a+c}{2} \right)$

$\displaystyle \frac{b+d}{2} = \left(\frac{a-c}{d-b}\right)\left(\frac{a+c}{2}\right)$

$\displaystyle b+d = \frac{(a-c)(a+c)}{d-b}$

$\displaystyle d^2 - b^2 = a^2 - c^2$

But where do you go from there?

Re: Prove the perpendicular bisector of a chord passes through center

Quote:

Originally Posted by

**Ragnarok** Okay, putting 0 in for x and y I get:

$\displaystyle 0 - \frac{b+d}{2} = \frac{a-c}{d-b} \left (0 - \frac{a+c}{2} \right)$

$\displaystyle \frac{b+d}{2} = \left(\frac{a-c}{d-b}\right)\left(\frac{a+c}{2}\right)$

$\displaystyle b+d = \frac{(a-c)(a+c)}{d-b}$

$\displaystyle d^2 - b^2 = a^2 - c^2$

But where do you go from there?

But you know that $\displaystyle a^2+b^2=r^2=c^2+d^2$.

Re: Prove the perpendicular bisector of a chord passes through center

Right. : ) Thank you so much for your help –it kills me when I get stuck on a problem and I have no idea who I could have asked. You rock!