Line joining vertex A of triangle ABC and orthocenter (H) meets the side BC in D. Then prove that
(a) `BD : DC = tan C : tan B`
(b) `AH : HD = (tan B + tan C) : tan A`

Explanation

In figure in `DeltaADB, BD = c cos B` (projection of AB on BC)
In `DeltaADC, CD = b cos C` (projection of AC on BC)
`:. (BD)/(CD) = (c cosB)/(b cos C) = (2 R sin C cos B)/(2R sin B cos C) = (tan C)/(tan B)`
Also, `(AH)/(HD) = (2R cos A)/(2R cos B cos C)`
`= (sin A)/(tan A cos B cos C)`
`= (sin (B + C))/(tan A cos B cos C)`
`= (sin B cos C + sin C cos B)/(tan A cos B cos C)`
`= (tan B + tan C)/(tan A)`