# Diameters and Circles

I have a question (given by a teacher) that looks really easy but then when I thought about it, couldn't find a way to find the answer. It is a proof question relating to diameters:

Prove that any two diameters are congruent, and that each diameter is twice as long as each radius.

It seems like common sense, but I can't think of a way to prove it.

Besides this, I also have another question (that I stumbled upon i my reference book) about circles that I could not figure out:

If one chord is a perpendicular bisector of a different chord, then the first chord is a diameter

I think there must be a property relating the center of a circle and the perpendicular bisector of any chord, but I am not sure where to start with this problem.

Any help or hints would be very nice :)

Thank you.

• Hint: Two diameters intersect each other on the center and then you can consider two triangles. Note the congruent angles opposite by the vertices. – Sigur Oct 28 '13 at 15:01
• I would think that it is the definition of "circle" that any two radii are congruent, and from there the facts about diameters follow. – hardmath Oct 28 '13 at 15:01
• @Sigur Doing that we can prove that the 2 triangles are congruent. What next? – Paul Filch Oct 28 '13 at 15:11
• @PaulFilch, they are congruent and isosceles since their sides are the circle radius. So you'll have congruent diameters. – Sigur Oct 28 '13 at 15:16
• @Sigur Oh, didn't realize they were isosceles. Any hints on the 2nd question – Paul Filch Oct 28 '13 at 15:25

Hint for first part: Consider two diameters on a circle, of length $a$ and $b$. Since a diameter is a chord of maximum length, diameter $b$ is greater than or equal to all other possible chord lengths, including $a$. Similarly diameter $a$ is greater than or equal tochord $b$. But $a\leq b$ and $b\leq a$ together imply $a=b$. Hence, any two diameters have the same length, hence any two diameters are congruent.

To show that this maximum chord length is $2r$, you need to show that chords which go through the center are longer than chords which don't.

• I think I got the first part thanks to Sigur. Any hints on the second question? – Paul Filch Oct 28 '13 at 15:26

The proof is as follows.

Join FA,FD,FB. As F is the center, FA=FB. Also, as D bisects AB, we have DA=DB. As FD is common, then from Triangle Side-Side-Side Equality, △ADF=△BDF.

In particular, ∠ADF=∠BDF; both are right angles

Thus, by definition, F lies on the perpendicular bisector of AB. Hence the result.