in triangle abc ,AD is an altitude and H is the orthocenter . prove that AH: HD= (tan B+ tan C) : Tan A
 

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$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \\ \mathrm{Distance} \mathrm{of} \mathrm{orthocentre} \mathrm{from} \mathrm{vertex} \mathrm{A} \mathrm{is} 2\mathrm{RcosA} \mathrm{and} \mathrm{from} \mathrm{side} \mathrm{BC} \\ \mathrm{is} 2\mathrm{RcosBcosC} \\ \mathrm{i}.\mathrm{e}. \\ \mathrm{AH}=2\mathrm{RcosA} \\ \mathrm{and} \\ \mathrm{HD}=2\mathrm{RcosBcosC} \\ \frac{\mathrm{AH}}{\mathrm{HD}}=\frac{2\mathrm{RcosA}}{2\mathrm{RcosBcosC}} \\ \frac{\mathrm{AH}}{\mathrm{HD}}=\frac{\mathrm{cosA}}{\mathrm{cosB}.\mathrm{cosC}}....\left(\mathrm{i}\right) \\ \mathrm{Now} \\ \frac{\mathrm{tanB}+\mathrm{tanC}}{\mathrm{tanA}}=\frac{{\displaystyle \frac{\mathrm{sinB}}{\mathrm{cosB}}}+{\displaystyle \frac{\mathrm{sinC}}{\mathrm{cosC}}}}{{\displaystyle \frac{\mathrm{sinA}}{\mathrm{cosA}}}} \\ =\mathrm{cosA}.\frac{\mathrm{sinB}.\mathrm{cosC}+\mathrm{cosB}.\mathrm{sinC}}{\mathrm{sinA}.\mathrm{cosB}.\mathrm{cosC}} \\ =\frac{\mathrm{cosA}.\sin \left(\mathrm{B}+\mathrm{C}\right)}{\mathrm{sinA}.\mathrm{cosB}.\mathrm{cosC}} \\ =\frac{\mathrm{cosA}.\sin \left(\mathrm{\pi }-\mathrm{A}\right)}{\mathrm{sinA}.\mathrm{cosB}.\mathrm{cosC}} \\ =\frac{\mathrm{cosA}.\mathrm{sinA}}{\mathrm{sinA}.\mathrm{cosB}.\mathrm{cosC}}=\frac{\mathrm{cosA}}{\mathrm{cosB}.\mathrm{cosC}} \\ \Rightarrow \frac{\mathrm{tanB}+\mathrm{tanC}}{\mathrm{tanA}}=\frac{\mathrm{cosA}}{\mathrm{cosB}.\mathrm{cosC}}....\left(\mathrm{ii}\right) \\ \mathrm{By} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \\ \frac{\mathrm{AH}}{\mathrm{HD}}=\frac{\mathrm{tanB}+\mathrm{tanC}}{\mathrm{tanA}} \left(\mathrm{Hence} \mathrm{Proved}\right) \end{gathered}$$

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