Let ABC be an acute angled triangle whose orthocentre is at H. If altitude from A is produced to meet the circumcircle of triangle ABC at `D` , then prove `H D=4RcosBcosC`

Explanation

In the figure, altitude AD meets BC at D and circumcircle at P. In circumcircle of triangle ABC, chord AB subtends same angle at point C and P.
`:. Angle BPA = angle BCA = C`
i.e., `angle BPD = C`
Also `angle HBD = 90^(@) -C :. angle BHD = C`
Thus `Delta BPD and DeltaBHD` are similar.
`:. HD = DP`
Therefore, P is image of H in BC
Also, `HP = 2HD = 2(2R cos B cos C)`
`= 4R cos B cos C`