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Let $CH -$ height in acute-angled triangle $ABC$. Some points $K$ and $N$ are on side $AB$. Let $O_1 -$ orthocenter of triangle $ACN$ and $O_2 -$ orthocenter of triangle $BCK$. Prove $$O_1=O_2=O \Leftrightarrow \frac{KH}{NH}=\frac{AH}{BH}$$

My work so far:

I solved this problem:

If $\angle C=90^{\circ}$. Then $$O_1=O_2=O \Leftrightarrow \frac{BN}{AK}=\tan^2A$$

Now I want to generalize this problem

enter image description here

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  • $\begingroup$ I can't understand whether the left hand side condition will ever hold true or not. Because, if that holds true i.e. say, $O_1 = O$, then as per your figure, $AO$ must be perpendicular to both $CN$ and $CB$, which is not possible unless $CN$ and $CB$ are parallel to each other....which further means that $N$ and $B$ must be the same point.... $\endgroup$
    – seavoyage
    Jul 14, 2016 at 18:32

1 Answer 1

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We will consider your problem from another perspective. First, we will consider only one triangle (let it be $\triangle ACN$), and then suppose we are given the line $AN$ (but not where $A$ and $N$ are) and points $C$ and $O$.

Now, the question is where $A$ and $N$ could be relatively to each other? It turns out, that $$|AH| \cdot |HN| = |OH| \cdot |HC| \tag{$\spadesuit$}.$$

Consider the following diagram:

pic

Wherever we place $A$, it also determines the line $AO$ which is one of the heights of the triangle. As the corresponding side $CN$ has to be perpendicular (the right angle marked by red square), it also determines the place where $N$ has to be, that is, how the whole triangle has to look like.

Observe that triangles $\triangle AHC$ and $\triangle OHN$ are similar, so $\frac{|AH|}{|HC|} = \frac{|OH|}{|HN|}$. Having $(\spadesuit)$ it is straightforward to prove thesis of your problem.

I hope this helps $\ddot\smile$

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