TITLE: Let $D,E,F$ be (respectively) the projections of $O$ on $BC,CA,AB$. Prove that $\cot{\angle ADB} + \cot{\angle BEC} + \cot{\angle CFA} =0$
QUESTION [6 upvotes]: 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 $$\cot{\measuredangle ADB} + \cot{\measuredangle BEC} + \cot{\measuredangle CFA} =0$$

REPLY [2 votes]: 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.