If in ` A B C ,` the distances of the vertices from the orthocentre are x, y, and z, then prove that `a/x+b/y+c/z=(a b c)/(x y z)`

Explanation

We know that distance of orthocenter (H) from vertex (A) is `2R cos A`
or `x = 2R cos A`
Similarly, `y = 2R cos B, z = 2R cos C`
`rArr (a)/(x) + (b)/(y) + (c)/(z) = (2R sin A)/(2R cos A) + (2R sin B)/(2R cos B) + (2R sin C)/(2 R cos C)`
`=tan A + tan B + tan C`
`= tan A tan B tan C`
Also, `(abc)/(xyz) = ((2R sin A) (2R sin B) (2R sin C))/(2R cos A ) (2R cos B) (2R cos C)`
`= tan A tan B tan C`
Hence, `(a)/(x) + (b)/(y) + (c)/(z) = (abc)/(xyz) ( :' tan A + tan B + tan C = tan A. tan B . tan C)`