Prove that $\frac{BC}{AH}×\frac{CA}{BH}×\frac{AB}{CH}$ = $\frac{BC}{AH}+\frac{CA}{BH}+\frac{AB}{CH}$, where $H$ is the orthocenter of $\triangle ABC$

Question

In $\triangle ABC$, three altitude lines $AE$, $BF$ and $CD$ dropped from three different vertex point intersect one another at point $H$ and $H$ is its orthocenter. Prove that $$\frac{BC}{AH}×\frac{CA}{BH}×\frac{AB}{CH} = \frac{BC}{AH}+\frac{CA}{BH}+\frac{AB}{CH}$$

Let express $AB=a$, $BC=b$ and $CA=c$. But how can I relate the length of $AH$, $BH$ and $CH$ with $a$, $b$ and $c$ or how can I express it? Is there another proof to solve this? I'd highly like to see that method.

EDIT: The diagram drawn above isn't satisfactory at all for proving the desired condition.

Answer 1

One approach? Trigonometry. If $A,B,C$ are the angles of the triangle at the vertices of the same name, then $\angle BHD = \angle FHC = A$, $\angle DHA=\angle CHE=B$, and $\angle AHF=\angle EHB=C$.

From this, $\frac{AD}{AH}=\sin B$, $\frac{DC}{AD}=\frac{\sin A}{\cos A}$, and $\frac{BC}{DC}=\frac{1}{\sin B}$. Multiply them, and $$\frac{BC}{AH}=\frac{AD}{AH}\cdot \frac{DC}{AD}\cdot\frac{BC}{DC}=\frac{\sin B}{1}\cdot\frac{\sin A}{\cos A}\cdot \frac{1}{\sin B}=\frac{\sin A}{\cos A}=\tan A$$ Similarly, $\frac{CA}{BH}=\tan B$ and $\frac{AB}{CH}=\tan C$.

Now, we wish to show that for angles $A,B,C$ of a triangle, $$\tan A+\tan B+\tan C = \tan A\tan B\tan C$$ Recall the angle addition formula for the tangent: $$0=\tan(A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan A\tan C-\tan B\tan C}$$ The tangent of the sum is zero because the sum is $180^\circ$, of course. That can only happen when the numerator is zero - and then $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ as desired.

OK, the fraction can also be zero if the denominator is infinite, which happens when one of $A,B,C$ is a right angle. In that case, $H$ coincides with one of the vertices and one of $AH,BH,CH$ is zero. We can claim that case as a success, with both sides of the equality equal to $\infty$.

Answer 2

I think everything is OK with this proof.

From the equation of angles we get the similarity of triangles. And from similarity of triangles we change the relations with tangents.

Only the last part may be more clear if we proof the formula

$\tan A +\tan B +\tan C = \tan A \tan B \tan C $

You can stop the proof here because from trigonometry we know, that if $A+B+C = \pi$ then this formula if true.

$\tan A+\tan B + \tan C = \tan A+\tan B + \tan (\pi - (A+B))= \tan A+\tan B - \tan(A+B)$

We also know, that

$\tan A + \tan B = \tan (A+B)(1-\tan A \tan B)$

So the above formula will be

$ \tan A+\tan B - \tan(A+B) = \tan (A+B)(1-\tan A \tan B) - \tan(A+B)= -\tan (A+B)\tan A \tan B= -\tan A \tan B \tan (\pi -C) =\tan A \tan B \tan C $

I think it is clear than claiming of equality of two infinities.