The discussion centers on proving that any chord perpendicular to the diameter of a circle is bisected by that diameter. Participants suggest using congruent triangles to demonstrate this property, specifically referencing the Side-Side-Side (SSS) congruence criterion. The key insight is that the intersection point of the chord and diameter, denoted as X, allows for the formation of two triangles that share a common segment and have equal hypotenuses, leading to congruence. However, it is clarified that the proof cannot rely on SSS without assuming the conclusion.
PREREQUISITESStudents studying geometry, mathematics educators, and anyone interested in understanding geometric proofs involving circles and chords.
What triangles are you forming ?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.
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).SammyS said:What triangles are you forming ?
Yes, the triangles are congruent, but not by SSS. That would require that you assume the thing you are to prove.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).