Thread: Proof: An Isosceles Triangle inscribed in a Circle

1. Proof: An Isosceles Triangle inscribed in a Circle

Hypothesis: AC = CB
DE is the diameter
CH is perpendicular to AB (the altitude)
Thesis: DE : BC = BC : CH (BC^2 = DE * CH)

2. Re: Proof: An Isosceles Triangle inscribed in a Circle

Originally Posted by goby

Hypothesis: AC = CB
DE is the diameter
CH is perpendicular to AB (the altitude)
Thesis: DE : BC = BC : CH (BC^2 = DE * CH)
$\displaystyle A=\frac{1}{2} *AB*CH$

$\displaystyle R=\frac{AB*BC*CA}{4A}$ (Circumradius - AoPSWiki)

$\displaystyle R=\frac{AB*BC*CA}{4*\frac{1}{2}*AB*CH}$

$\displaystyle R=\frac{BC*CA}{2CH}$

$\displaystyle R=\frac{BC^2}{2CH}$

$\displaystyle \frac{BC^2}{CH}=2R=DE$

$\displaystyle BC^2=DE*CH$

3. Re: Proof: An Isosceles Triangle inscribed in a Circle

Okay, I understood it, but what's the proof for R = ?

4. Re: Proof: An Isosceles Triangle inscribed in a Circle

Originally Posted by goby
Okay, I understood it, but what's the proof for R = ?
Area of a Triangle in Terms of Circumradius - ProofWiki

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prove that the triangle is isosceles which is inscribe in a circle

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