Let $H$ be the orthocenter of a triangle $ABC$, prove that $HA+HB+HC=a\cot A+b\cot B+c\cot C$

Question

Let $H$ be the orthocenter of a triangle $ABC$, prove that in any triangle $ABC$ the relation holds true: $$HA+HB+HC=a\cot A+b\cot B+c\cot C.$$ I found this problem in my 9th grade textbook and I have tried to change the form of: $a\cot A+b\cot B+c\cot C = 2(R+r)$, but this doesn't seem to solve it leaving me clueless. Can anyone help me with an idea that I can further continue to successfully solve this problem?

Answer 1

There are 2 cases to consider.

Case (i), when $\Delta ABC$ is an acute angled triangle.

In the figure, simple angle chasing shows that $\angle ABH=90^o-A$ and $\angle BHA=180^o-C$.

Applying Sine Rule on $\Delta ABH$

$$\frac{AH}{\sin (90^o-A)}=\frac{AB}{\sin (180^o-C)}$$

$$AH=\frac{c \cos A}{\sin C}$$

$$AH=\frac{a \cos A}{\sin A}= a \cot A$$

Similarly we can prove that $BH=b \cot B$ and $CH=c \cot C$.

Case (ii), when $\Delta ABC$ is an obtuse angled triangle. Let's say $\angle A$ is obtuse.

Using the facts that $AEHF$ and $BCEF$ are concyclic, we can prove that $\angle AHB=C$, $\angle HBA=A-90^o$ and $\angle BAH=B+90^o$

Applying Sine Rule on $\Delta HAB$, we have

$$\frac{BH}{\sin (B+90^o)}=\frac{AB}{\sin C}=\frac{HA}{\sin (A-90^o)}.$$

From $$\frac{BH}{\sin (B+90^o)}=\frac{AB}{\sin C},$$ we have $$HB=\frac{c \cos B}{\sin C}=\frac{b \cos B}{\sin B}= b \cot B.$$

This is the same as the acute angled case.

But from $$\frac{AB}{\sin C}=\frac{HA}{\sin (A-90^o)},$$ we have

$$HA=\frac{c (-\cos A) }{\sin C}=\frac{a (-\cos A) }{\sin A}=-a \cot A$$

Note the existence of the minus sign when $A$ is obtuse.

Thus when $A$ is obtuse, the result should be:

$$HA+HB+HC=-a\cot A+b\cot B+c\cot C$$

Or more generally, the result should be written as: $$HA+HB+HC=a\lvert \cot A \rvert+b\lvert \cot B \rvert+c\lvert \cot C \rvert$$