Circle Geometry Proof: Perpendicular Chord Bisected by Diameter

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Mr Davis 97
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Homework Statement


Prove that any chord perpendicular to the diameter of a circle is bisected by the diameter.

Homework Equations

The Attempt at a Solution


I was thinking that maybe I could form two triangles, show that these triangles are congruent, and then conclude that the two lengths of the chord cut by the diameter are equal in length. But I can't seem to prove congruence.
 
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Mr Davis 97 said:

Homework Statement


Prove that any chord perpendicular to the diameter of a circle is bisected by the diameter.

Homework Equations

The Attempt at a Solution


I was thinking that maybe I could form two triangles, show that these triangles are congruent, and then conclude that the two lengths of the chord cut by the diameter are equal in length. But I can't seem to prove congruence.
What triangles are you forming ?
 
SammyS said:
What triangles are you forming ?
Oh wait... Let X be the intersection of the chord and the diameter. If I form triangles with the radius, then I get that the hypotenuses are equal, but I also get that the segment from X to the center of the circle is the same for both triangles, so they are congruent by SSS (since the other side for both triangles comes from the Pythagorean theorem).
 
Mr Davis 97 said:
Oh wait... Let X be the intersection of the chord and the diameter. If I form triangles with the radius, then I get that the hypotenuses are equal, but I also get that the segment from X to the center of the circle is the same for both triangles, so they are congruent by SSS (since the other side for both triangles comes from the Pythagorean theorem).
Yes, the triangles are congruent, but not by SSS. That would require that you assume the thing you are to prove.