Proof : How to find the radius of the circumcircle of a triangle | Triangle inside a circle
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- เผยแพร่เมื่อ 13 ก.ย. 2024
- Proof : How to find the radius of the circumcircle of a triangle | Triangle inside a circle
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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
In fact proof for obtuse angle is analogous but picture is slightly different
Proof for right angle is different
We have right angle as inscribed angle so subtended arc is semicircle so hypotenuse is diameter
An essential tool for mathletes!!Kudos.
Yes!!Well done. How about an obtuse triangle? Does the formula hold?
By considering area formula of two adjacent sides and their angle with double angle at center,.......
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 *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.
Nice elaboration.......
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
Why such long solution.
Area =bxcsinQ/2
Sin Q=a/2R
So area=1/2bxcxa/2R
A=abc\4R
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 --- sin(A)/a=sin(B)/b=sin(C)/c=1/2R ( 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 .)
Yes , for law of sines we need these three cases which i mentioned , for formula for radius not necessarily because each non degenerate triangle has at least two acute angles
Technically you are correct . There is another derivation of the law of sines , the one where the area of the triangle is equal to A = 1/2(base)(height) = 1/2a(bsinC)=1/2a(csinB) . Divide though by 1/2abc , and you get : sinC/c=sinB/b . We also have that the area of the triangle is A=1/2(base)(height)=1/2c(asinB)=1/2c(bsinA).Now , divide this through by 1/2abc , and you get sinB/b=sinA/a. We therefore have sinA/a=sinB/b=sinC/c . Now , this derivation depends on two heights from two different bases . These heights are always positive even if one of the three angles is abtuse . This derivation requires no circle ; but , if a circumscribing circle is drawn of radius R , we get sinA/a=sinB/b=sinC/c=1/(2R) because if one of the angles is 90 degrees ; then , sin90=1 and the opposite side = 2R . The diameter of the circle .@@holyshit922
Thanks
asnwer=76 isit