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(A) Let $\Delta$ be the area of the triangle. The perpendicular distances from the circumcenter to the sides $a, b, c$ are $x = R \cos A$,$y = R \cos

Vedclass · Accepted Answer

$\frac{a^2 + b^2 + c^2}{2R}$

Vedclass · Answer

(A) Let $\Delta$ be the area of the triangle. The perpendicular distances from the circumcenter to the sides $a, b, c$ are $x = R \cos A$,$y = R \cos B$,and $z = R \cos C$.However,the problem states $x, y, z$ are perpendiculars to sides $a, b, c$. In a triangle,the perpendicular distance from the circumcenter to side $a$ is $R \cos A$. Given $\Delta = \frac{1}{2}ax = \frac{1}{2}by = \frac{1}{2}cz$,we have $x = \frac{2\Delta}{a}$,$y = \frac{2\Delta}{b}$,and $z = \frac{2\Delta}{c}$.Substituting these into the expression: $\frac{bx}{c} + \frac{cy}{a} + \frac{az}{b} = \frac{b}{c}(\frac{2\Delta}{a}) + \frac{c}{a}(\frac{2\Delta}{b}) + \frac{a}{b}(\frac{2\Delta}{c})$$= 2\Delta \left( \frac{b^2 + c^2 + a^2}{abc} \right)$Using $\Delta = \frac{abc}{4R}$,we get:$= 2(\frac{abc}{4R}) \left( \frac{a^2 + b^2 + c^2}{abc} \right) = \frac{a^2 + b^2 + c^2}{2R}$.

List-$I$	List-$II$
$(A) \ r_1 r_2 \sqrt{\frac{4R-r_1-r_2}{r_1+r_2}}$	$1. \ b$
$(B) \ \frac{r_2(r_3+r_1)}{\sqrt{r_1r_2+r_2r_3+r_3r_1}}$	$2. \ a^2, b^2, c^2 \text{ are in } AP$
$(C) \ \frac{a}{c} = \frac{\sin(A-B)}{\sin(B-C)}$	$3. \ \Delta$
$(D) \ bc \cos^2 \frac{A}{2}$	$4. \ R r_1 r_2 r_3$
	$5. \ s(s-a)$

If $x, y, z$ are the lengths of the perpendiculars drawn from the circumcenter to the sides $a, b, c$ respectively of a triangle,then the value of $\frac{bx}{c} + \frac{cy}{a} + \frac{az}{b}$ is

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