Proof : How to find the radius of the circumcircle of a triangle | Triangle inside a circle

Proof : How to find the radius of the circumcircle of a triangle | Triangle inside a circle

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[prime]

An essential tool for mathletes!!Kudos.

[holyshit]

Maybe we should consider three cases
1) theta is acute angle
2) theta is right angle
3) theta is obtuse angle
In this video only case 1) is proven

[prime]

Yes!!Well done. How about an obtuse triangle? Does the formula hold?

[Irtsak]

In Greece, in the textbook of the secondary school there is the following simple proof.
I have taught it over 1000 times ! 😊
Let triangle ABC inscribed in a circle of radius R.
Draw the height *AE=h* of the triangle and the diameter *AD=2R*
Obviously orthogonal triangles ABE and ADC are similar cause <ABE=<ADC (inscribed angles meeting the same arc AC)
So AB/AD=AE/AC => c/2R=h/b => *h=(b⋅c)/2R* (1)
Area of triangle ABC = 1/2⋅ BC⋅AE=1/2 a⋅h=1/2⋅(a⋅b⋅c)/2R=abc/4R cause (1)
So *ο.ε.δ* (όπερ έδει δείξαι in Greek) or *Q.E.D* ( Quod Erat Demonstrandum in Latin) means :
*“I proved what you asked me to do”*
and is found at the end of some simple, impressive and at the same time visually appealing proofs. In one sense, it is synonymous with truth and beauty in Mathematics.

[misterenter-izrz]

By considering area formula of two adjacent sides and their angle with double angle at

[isg]

Why such long solution.
Area =bxcsinQ/2
Sin Q=a/2R
So area=1/2bxcxa/2R
A=abc\4R

[solimana-soli]

I knew a different proof, with similarity between right triangles ABH (with AH height of ABC) and ACE ( with AE diameter as in the previous post)
AB : AH = AE : AC
c : AH = 2r : b
knowing that AH = 2*area/a (area = A)
(2*A/a)*2r = bc and then
r = a*b*c/4*A

[pk]

This is a very nice simple proof ; but, we must not forget that it also proves the law of sines because he could have chose any of the three sides . Thus --- ( I am using A as the angle at A, B as the angle at B, and C as the angle at C because my keyboard does not have greek letters .)