as shown in the figure AD is the altitude on BC and AD produced meets the circumcircle of traingle ABC at P where DP =x . similarly EQ=y and FR = z . if a, b ,c respectively denotes the sides BC, CA, and AB , then a/2x+b/2y+c/2z has the value equal to

Aarushi Ahlawat
41 Points
6 years ago
Use the intersection chord theorem to get relation between a/2x, b/2y and c/2z. NOw write a, b, and c as sum of two parts of the side e.g a=BD+DC and so on. Convert the ratios to trigonometric ratios of specific angle and you should get answer in the form of sum of trigonometric ratios of the angle angles. Further simplication is possible with the fact that A+B+C=pi.

for Two intersecting chords BC and AP

$AD X x=BD X DC$

$\frac{a}{2x}=\frac{(BD+ DC)AD}{2BDXDC}$

$=\frac{1}{2}\left ( \frac{AD}{DC}+\frac{AD}{BD} \right )$
$=\frac{1}{2}\left ( tan(C)+tan(B) \right )$

similarly

$\frac{b}{2y}=\frac{1}{2}\left ( tan(A)+tan(C) \right )$
$\frac{c}{2z}=\frac{1}{2}\left ( tan(A)+tan(B) \right )$

Hence

$\frac{a}{2x}+\frac{b}{2y}+\frac{c}{2z}=tan(A)+tan(B)+tan(C)$