A right angled triangle circumscribes a circle with radius r . prove that the half of sum of all sides of triangle is equal to radius ( AB+BC+CA/2 = r )
Answers
Step-by-step explanation:
Let $a, b$ and $c$ denote the triangle's three sides and let $A$ denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply $R=\frac{abc}{4A}$. This can be rewritten as $A=\frac{abc}{4R}$.
Proof
[asy] pair O, A, B, C, D; O=(0,0); A=(-5,1); B=(1,5); C=(5,1); dot(O); dot (A); dot (B); dot (C); draw(circle(O, sqrt(26))); draw(A--B--C--cycle); D=-B; dot (D); draw(B--D--A); label("$A$", A, W); label("$B$", B, N); label("$C$", C, E); label("$D$", D, S); label("$O$", O, W); pair E; E=foot(B,A,C); draw(B--E); dot(E); label("$E$", E, S); draw(rightanglemark(B,A,D,20)); draw(rightanglemark(B,E,C,20)); [/asy]
We let $AB=c$, $BC=a$, $AC=b$, $BE=h$, and $BO=R$. We know that $\angle BAD$ is a right angle because $BD$ is the diameter. Also, $\angle ADB = \angle BCA$ because they both subtend arc $AB$. Therefore, $\triangle BAD \sim \triangle BEC$ by AA similarity, so we have\[\frac{BD}{BA} = \frac{BC}{BE},\]or\[\frac {2R} c = \frac ah.\]However, remember that $[ABC] = \frac {bh} 2\implies h=\frac{2 \times [ABC]}b$. Substituting this in gives us\[\frac {2R} c = \frac a{\frac{2 \times [ABC]}b},\]and then simplifying to get\[R=\frac{abc}{4\times [ABC]}\text{ or }[ABC]=\frac{abc}{4R}\]and we are done.
Formula for Circumradius
$R = \frac{abc}{4rs}$ Where $R$ is the circumradius, $r$ is the inradius, and $a$, $b$, and $c$ are the respective sides of the triangle and $s = (a+b+c)/2$ is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that $A = rs$.
But, if you don't know the inradius, you can find the area of the triangle by Heron's Formula:
$A=\sqrt{s(s-a)(s-b)(s-c)}$
Euler's Theorem for a Triangle
Let $\triangle ABC$ have circumcenter $O$ and incenter $I$.Then\[OI^2=R(R-2r) \implies R \geq 2r\]
Proof