Let O be the circumcentre and H be the orthocentre of an acute angled triangle ABC. If `A gt B gt C`, then show that `Ar (Delta BOH) = Ar (Delta AOH) + Ar (Delta COH)`

Explanation

From the figure, we have
`angle OAH = A - 2(90^(@) - B) = B - C`
Similarly, `angle OBH = A - C`
and `angle OCH = A - B`
Also, `AH = 2R cos A, BH = 2R cos B, CH = 2R cos C`
`Ar (Delta AOH) = (1)/(2) (R) (2R cos A) sin (B - C)`
`= R^(2) cos (B + C) sin (C - B)`
`=(R^(2))/(2) (sin 2 C - sin 2B)`
Similarly, `Ar(DeltaBOH) = (R^(2))/(2) (sin 2C - sin 2A)`
and `Ar (Delta COH) = (R^(2))/(2) (sin 2 B - sin 2A)`
Clearly, `AR (Delta AOH) + Ar (Delta COH) = Ar (Delta BOH)`