Question

If $P_1,P_2,P_3$ are lengths of perpendicular from vertices A, B, C respectively of a $△ABC$, then prove that
$R=\frac{1}{4}\frac{a^2+b^2+c^2}{P_{1}cosA+P_{2}cosB+P_{3}cosC}$

Answer 1

Let $\Delta$ be the area of triangle ABC.
Then $\Delta =\frac{1}{2}{P}_{1}a=\frac{1}{2}{P}_{2}b=\frac{1}{2}{P}_{3}c$
$\Rightarrow P_1=\frac{2\Delta }{a};P_2=\frac{2\Delta }{b};P_3=\frac{2\Delta }{c}$
Consider,$\frac{1}{4}\frac{a^2+b^2+c^2}{P_{1}cosA+P_{2}cosB+P_{3}cosC}$
$=\frac{1}{4}\frac{a^2+b^2+c^2}{\frac{2\Delta }{a}\left(\frac{b^2+c^2-a^2}{2bc}\right)+\frac{2\Delta }{b}\left(\frac{a^2+c^2-b^2}{2ac}\right)+\frac{2\Delta }{c}\left(\frac{a^2+b^2-c^2}{2ab}\right)}$
$=\frac{1}{4}\frac{a^2+b^2+c^2}{\frac{\Delta }{abc}(a^2+b^2+c^2)}$
$=\frac{abc}{4\Delta }$
$=R$