Circle Geometry Proof: Perpendicular Chord Bisected by Diameter

AI Thread Summary
A proof is sought to demonstrate that any chord perpendicular to a circle's diameter is bisected by that diameter. The initial approach involves forming two triangles to show congruence, but the challenge lies in proving this congruence. The discussion reveals that the intersection point of the chord and diameter can be used to form triangles with the radius, leading to the conclusion that they are congruent by SSS. However, it is pointed out that using SSS for this proof is flawed, as it assumes the conclusion that needs to be proven. Clarifying the correct method for establishing the congruence of the triangles is essential for completing the proof.
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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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