Question
If $$x, y$$ and $$z$$ are perpendiculars drawn on $$a, b$$ and $$c,$$ respectively, then the value of $$\frac{{bx}}{c} + \frac{{cy}}{a} + \frac{{az}}{b}$$ will be
A.
$$\frac{{{a^2} + {b^2} + {c^2}}}{{2R}}$$
B.
$$\frac{{{a^2} + {b^2} + {c^2}}}{{R}}$$
C.
$$\frac{{{a^2} + {b^2} + {c^2}}}{{4R}}$$
D.
$$\frac{{2\left( {{a^2} + {b^2} + {c^2}} \right)}}{R}$$
Answer :
$$\frac{{{a^2} + {b^2} + {c^2}}}{{2R}}$$
Solution :
Let area of triangle be $$\Delta ,$$ then according to question,
$$\eqalign{
& \Delta = \frac{1}{2}ax = \frac{1}{2}by = \frac{1}{2}cz \cr
& \therefore \frac{{bx}}{c} + \frac{{cy}}{a} + \frac{{az}}{b} \cr
& = \frac{b}{c}\left( {\frac{{2\Delta }}{a}} \right) + \frac{c}{a}\left( {\frac{{2\Delta }}{b}} \right) + \frac{a}{b}\left( {\frac{{2\Delta }}{c}} \right) \cr
& = \frac{{2\Delta \left( {{b^2} + {c^2} + {a^2}} \right)}}{{abc}} \cr
& = \frac{{2\left( {{a^2} + {b^2} + {c^2}} \right)}}{{abc}} \cdot \frac{{abc}}{{4R}} = \frac{{{a^2} + {b^2} + {c^2}}}{{2R}} \cr} $$