$\tan B\cdot \frac{BM}{MA}+\tan C\cdot \frac{CN}{NA}=\tan A. $ Let $\triangle ABC$ be a triangle and $H$ be the orthocenter of the triangle. If $M\in AB$ and $N \in AC$ such that $M,N,H$ are collinear prove that : 
$$\tan B\cdot \frac{BM}{MA}+\tan C\cdot \frac{CN}{NA}=\tan A. $$
Thanks :) 
 A: 
with menelaus' theorem, 
in$\triangle ABD$,$\dfrac{BM}{MA}\dfrac{AH}{HD}\dfrac{DK}{KB}=1 $,ie, $\dfrac{BM}{MA}=\dfrac{BK*HD}{AH*DK}$. 
in$\triangle ACD$,$\dfrac{CN}{NA}\dfrac{AH}{HD}\dfrac{DK}{KC}=1 $,ie, $\dfrac{CN}{NA}=\dfrac{CK*HD}{AH*DK}$.
$tanB=\dfrac{AD}{BD}, tanC=\dfrac{AD}{DC},tanA=\dfrac{BC}{AH}$
LHS=$\dfrac{AD}{BD}*\dfrac{BK*HD}{AH*DK}+\dfrac{AD}{DC}*\dfrac{CK*HD}{AH*DK}$=$\dfrac{AD*HD}{AH*DK}(\dfrac{BK}{BD}+\dfrac{CK}{DC})$
$\dfrac{BK}{BD}+\dfrac{CK}{DC}=\dfrac{BK}{BD}+\dfrac{KB+BD+DC}{DC}=\dfrac{BK}{BD}+\dfrac{BK}{DC}+\dfrac{BD}{DC}+\dfrac{DC}{DC}=\dfrac{BK*BC}{BD*DC}+\dfrac{BC}{DC}=\dfrac{BC}{DC}*\dfrac{BK+BD}{BD}=\dfrac{BC*DK}{DC*BD}$
LHS=$\dfrac{AD*HD}{AH*DK}*\dfrac{BC*DK}{DC*BD}=\dfrac{BC*AD*HD}{AH*DC*BD}$
clearly, $\triangle BHD \sim \triangle ACD$,that is $\dfrac{BD}{AD}=\dfrac{HD}{CD}$, ie. $\dfrac{AD*HD}{DC*BD}=1$
LHS=$\dfrac{BC}{AH}=tanA$
for $tanA=\dfrac{BC}{AH}$, we can proof as follow:
$A=\angle BAD+\angle DAC=\angle HCD+\angle HBD$, so $tanA=tan(\angle HBD+\angle HCD)=\dfrac{tan\angle HBD+tan\angle HCD}{1-tan\angle HBD*tan\angle HCD}=\dfrac{\dfrac{HD}{BD}+\dfrac{HD}{DC}}{1-\dfrac{HD*HD}{BD*DC}}=\dfrac{HD*BC}{BD*DC-HD^2}=\dfrac{HD*BC}{AD*HD-HD^2}=\dfrac{HD*BC}{HD(AD-HD)}=\dfrac{BC}{AH}$
