Let $D,E,F$ be (respectively) the projections of $O$ on $BC,CA,AB$. Prove that $\cot{\angle ADB} + \cot{\angle BEC} + \cot{\angle CFA} =0$ 
Let $O$ be an arbitrary point located inside the triangle $ABC$. Let $D, E, F$ be (respectively) the projections of $O$ on $BC, CA, AB$. Prove that $$\cot{\measuredangle ADB} + \cot{\measuredangle BEC} + \cot{\measuredangle CFA} =0$$
 A: This is a nice question. I am assuming all the angles are directed. (Directed angles modulo $180^\circ$ are sufficient here, since $\cot$ is periodic with period $180^\circ$.)
We direct our three sidelines $BC$, $CA$ and $AB$. All distances along these lines will thus be understood to mean directed distances. Let $X$, $Y$, $Z$ be the feet of the altitudes of triangle $ABC$ emanating from $A$, $B$, $C$. Let $U$, $V$ and $W$ be the midpoints of the segments $BC$, $CA$, $AB$.
1. Pythagoras in the right-angled triangles $ODB$ and $ODC$ yields $BO^2 = BD^2 + OD^2$ and $CO^2 = CD^2 + OD^2$. Subtracting these two equalities yields $BO^2 - CO^2 = \left(BD^2 + OD^2\right) - \left(CD^2 + OD^2\right) = BD^2 - CD^2$. Similarly, $CO^2 - AO^2 = CE^2 - AE^2$ and $AO^2 - BO^2 = AF^2 - BF^2$. Summing the last three equalities, we obtain
$\left(BO^2 - CO^2\right) + \left(CO^2 - AO^2\right) + \left(AO^2 - BO^2\right) = \left(BD^2 - CD^2\right) + \left(CE^2 - AE^2\right) + \left(AF^2 - BF^2\right)$.
The left hand side of this equality simplifies to $0$, and so it becomes
(1) $0 = \left(BD^2 - CD^2\right) + \left(CE^2 - AE^2\right) + \left(AF^2 - BF^2\right)$.
This is nothing new so far (just a well-known equality, I think due to Carnot, which can be regarded as an additive analogue of Ceva's theorem, or at least a direction of it).
2. Since $U$ is the midpoint of $BC$, we have $BU = - CU$ (remember that our lengths are directed), so that $BU + CU = 0$. Now, adding the equalities $BD = BU + UD$ and $CD = CU + UD$, we obtain $BD + CD = \left(BU + UD\right) + \left(CU + UD\right) = \underbrace{BU + CU}_{=0} + 2 \cdot UD = 2 \cdot UD$. Now,
$BD^2 - CD^2 = \underbrace{\left(BD + CD\right)}_{=2 \cdot UD} \underbrace{\left(BD - CD\right)}_{= BC} = 2 \cdot UD \cdot BC$.
Similarly, $CE^2 - AE^2 = 2 \cdot VE \cdot CA$ and $AF^2 - BF^2 = 2 \cdot WF \cdot AB$. Using the last three equalities, we can rewrite (1) as
$0 = 2 \cdot UD \cdot BC + 2 \cdot VE \cdot CA + 2 \cdot WF \cdot AB$.
Dividing this by $2$, we obtain
(2) $0 = UD \cdot BC + VE \cdot CA + WF \cdot AB$.
3. Let $H$ be the orthocenter of triangle $ABC$. Then, $X$, $Y$, $Z$ are the feet of the perpendiculars from $H$ to $BC$, $CA$, $AB$. Thus, we can apply (2) to $H$, $X$, $Y$ and $Z$ instead of $O$, $D$, $E$ and $F$, respectively. We thus obtain
$0 = UX \cdot BC + VY \cdot CA + WZ \cdot AB$.
Subtracting this equality from (2), we obtain
$0 = \left(UD \cdot BC + VE \cdot CA + WF \cdot AB\right) - \left(UX \cdot BC + VY \cdot CA + WZ \cdot AB\right)$
$= \underbrace{\left(UD - UX\right)}_{= XD} \cdot BC + \underbrace{\left(VE - VY\right)}_{= YE} \cdot CA + \underbrace{\left(WF - WZ\right)}_{= ZF} \cdot AB$
(3) $= XD \cdot BC + YE \cdot CA + ZF \cdot AB$.
4. Let us direct the line $AX$ in such a way that rotating a positive-length vector on the line $BC$ by $90^\circ$ clockwise produces a positive-length vector on the line $AX$. Let us direct the lines $BY$ and $CZ$ similarly. Let $\Delta$ be the signed area of triangle $ABC$. Recall that the area of a triangle equals half times an altitude times the corresponding sidelength. Hence, $\Delta = \dfrac{1}{2} \cdot AX \cdot BC = \dfrac{1}{2} \cdot BY \cdot CA = \dfrac{1}{2} \cdot CZ \cdot AB$. (This is a honest equality, not just an equality that holds up to sign; and the reason for this is how we directed the lines $AX$, $BY$ and $CZ$.) From $\Delta = \dfrac{1}{2} \cdot AX \cdot BC$, we obtain $BC = \dfrac{2 \Delta}{AX}$. Similarly, $CA = \dfrac{2 \Delta}{BY}$ and $AB = \dfrac{2 \Delta}{CZ}$. Using the last three equalities, we can rewrite (3) as
$0 = XD \cdot \dfrac{2 \Delta}{AX} + YE \cdot \dfrac{2 \Delta}{BY} + ZF \cdot \dfrac{2 \Delta}{CZ}$.
Dividing this equality by $2 \Delta$ and simplifying, we obtain
(4) $0 = \dfrac{XD}{AX} + \dfrac{YE}{BY} + \dfrac{ZF}{CZ}$.
Now, the right-angled triangle $AXD$ yields $\dfrac{XD}{AX} = \cot \measuredangle ADX = \cot \measuredangle ADB$. (We are using the interplay between directed lengths and directed angles here -- again, the way how we directed the line $AX$ is important here.) Similarly, $\dfrac{YE}{BY} = \cot \measuredangle BEC$ and $\dfrac{ZF}{CZ} = \cot \measuredangle CFA$. Using the last three equalities, we can rewrite (4) as
$0 = \cot \measuredangle ADB + \cot \measuredangle BEC + \cot \measuredangle CFA$.
Done!
A: Let $\measuredangle ADB=\alpha$ and $AD', BE'$ be altitudes. Take $BD'=x$, $DD'=y$, $CD=z$. Assume $x+y+z=a$, (if $a=x+z-y$, we can follow similarly).
If $h_a$ is the altitude of $\triangle ABC$ on $BC$, we know that $2h_a (\cot B +\cot C)=a$. Using it on $\triangle ABD$ and $\triangle ACD$ gives:
$$\frac{\cot B +\cot\alpha}{\cot C-\cot\alpha}=\frac{BD}{CD}$$
On simplifying, we have: $$\frac{BD\cot C - CD\cot B}{a}=\cot\alpha$$
Thus, all we need to prove is:
$$\sum\limits_{cyc}BDbc\cot C - CDbc\cot B=0$$
$$\sum\limits_{cyc}BDb\cos C - CDc\cos B=0$$
$$\sum\limits_{cyc}(x+y)(z+y) - xz=ay=0$$
This follows easily from Carnot's theorem:
$$0=\sum_{cyc} (BD^2-CD^2)-(BD'^2-CD'^2)=(x+y)^2-x^2+(y+z)^2-z^2=2ay$$
Hence, we showed that $\sum_{cyc} ay=0$, and we are done. Note that if say $AC= CE'+AE-EE'$, then $AC\cdot EE'$, would have a negative sign, which is why the cyclic sum works.
