Proof: An Isosceles Triangle inscribed in a Circle

**goby** · September 6th 2011, 12:21 AM

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

**alexmahone** · September 6th 2011, 12:46 AM

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)

$A=\frac{1}{2} *AB*CH$

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

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

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

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

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

$BC^2=DE*CH$

**goby** · September 6th 2011, 01:19 AM

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

?

**alexmahone** · September 6th 2011, 01:25 AM

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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