# Minimalistic Proof for Existence of Orthocenter

It is a well known fact that the altitudes of a triangle $$ABC$$ ( with vertices $$A,B,C$$ ) intersect at exactly one point, the orthocenter $$H$$.

My question is if it is possible to finish the proof conducted by following concrete strategy (...I'm aware of that there are plenty different proofs of this fact, the concern I'm aiming to address in this question is really only about effectiveness of the presented strategy below):

One forms altitudes to side $$b$$ through through $$B$$ intersecting $$b$$ in $$B'$$ and altitude to $$c$$ through $$C$$ intersecting $$c$$ in $$C'$$. These intersect in $$H$$, our a priori potential candidate to be the orthocenter.
Then we consider the line $$AH$$ intersection side $$a$$ in $$A'$$. Note, that a priori we not know (actually, that's what we want to show) that $$AH$$ is perpendicular to $$a$$. See the auxiliary picture:

So a priori we not know that $$\delta$$ and $$\zeta$$ are $$90^{\circ}$$ and want to show this.

Now the actual problem/Question: Is it possible from this "ansatz" to show that $$AH$$ is perpendicular to $$a$$ - equivalently that $$\delta,\zeta =90^{\circ}$$ only by means of "juggling" with properties of similar triangles of the drawn subtriangles in the picture and using primitive angle chase?

So to keep the proof as "minimalistic" as possible in the way that the only primitive tools we can are actually allowed to use are just to exploit properties of similar triangles & basic triangle chase?

Especially without drawing additional "auxiliary" constructions like eg in the two proofs presented here or invoking additional auxiliary results, eg Ceva's Theorem, transferring problem to uniqueness of bisector of a auxiliary triangle, using properties of cyclic quadrilaterals.
Really, purely only using primitive similar triangle properties & basic triangle chase between drawn subtriangles.

The latter restriction not to use properties of cyclic quadrilaterals seems to be strange because at all this is not a deep and very useful auxiliary tool, but I wondering if it is even possible to give a proof even without it.

What is immediate so far from the construction is that following triangle pairs are similar: eg $$AC'C$$ & $$B'CH$$, $$AB'B$$ & $$C'HB$$, $$C'HB$$ & $$B'CH$$. Presumably there are some more which one could read up immediately but I not see more off the bat.
That appears that we are quite close to what the want but for example starting with $$\delta$$ and chasing through the relations between drawn angles the jungle of similar triangles allow so far one finishes finally instead with desired $$\delta=...= \zeta$$ as we want in something useless tautological as $$\delta=.... 180- \zeta$$ whatever one tries (...at least what ever I tried so far by means of the methods I intend to allow)

Here is a solution which is pretty close to what I want ... it uses exactly the expanse of methods I would like to allow to apply but it approaches the problem "from the other side":
It starts from a different auxiliary line $$A'H$$ assuming it to be perpendicular to $$a$$ and actually deduces with exactly the methods I would like to use that this is actually this line is actually the altitude through $$A$$.

One can also change the starting assumption to start with different auxiliary line, the foot/altitude at $$a$$ through $$A$$ intersecting (by assumption perpendicularly) $$a$$ in $$A'$$ and show with the methods I would prefer above that it actually must go through $$H$$.

So these both approached work similarly: We draw altitudes of $$(b,B)$$ and $$(c,C)$$ (that's coincides with the what I want to do above), but as additional auxiliary line one approach adds $$A'H$$ assuming as perpendicular line to $$a$$, and the other $$AA'$$ and reasons in the way I would like to do, ie only by means of similarities of triangles + angle chase.

Now, what I want is just to start with different auxiliary line, namely instead with auxiliary line $$AH$$ about which I a priori not know if it is perpendicular to $$a$$ and use similar techniques.

Now the question is can one see that it's possible to argue that way to conclude that $$AH$$ is perpend to $$a$$ (and so $$H$$ welldefined orthocenter), or is it possible to show that under this assumption and allowing only these primitive tools it actually not suffice to conclude it in that minimalistic way? Ie as above, at least some additional result, eg properties of cyclic quadrilaterals must be exploited.
So the focus of this question is the (non) feasibility of described strategy under retaining the radical minimalism on which tool I want actually use.

The question arised as specification of this one.

• Does this argument work for you? $$\frac{|AC'|}{|HC'|}=\frac{|AC'|}{|CC'|}\cdot \frac{|CC'|}{|HC'|} =\frac{|HC'|}{|BC'|}\cdot\frac{|CC'|}{|HC'|}=\frac{|CC'|}{|BC'|}$$ where the middle equality follows from $\triangle AC'C\sim \triangle HC'B$. Consequently, $\triangle CBC'\sim\triangle AHC'$, so that $\angle B=\angle AHC'=\angle A'HC$. Thus, two angles of $\triangle A'HC$ match two angles of $\triangle C'BC$, so that the third angles must match, too: $\angle HA'C$ is right.
– Blue
Commented Aug 1 at 1:42
• @That's perfect, exactly this kind of argument I was looking for, thanks! Commented Aug 1 at 11:27

We have $$\triangle ABC$$ with altitudes $$BB'$$ and $$CC'$$ meeting at $$H$$, and line $$AH$$ meeting $$BC$$ (not-necessarily-perpendicularly) at $$A'$$.
An easy angle chase shows that $$\angle A\cong\angle BHC'$$, so that right triangles $$\triangle AC'C$$ and $$\triangle HC'B$$ are similar. We can write $$\frac{|AC'|}{|HC'|}\quad=\quad\color{blue}{\frac{|AC'|}{|CC'|}}\cdot \frac{|CC'|}{|HC'|} \underbrace{\qquad=\qquad}_{\triangle AC'C\sim\triangle HC'B}\color{blue}{\frac{|HC'|}{|BC'|}}\cdot\frac{|CC'|}{|HC'|}\quad=\quad\frac{|CC'|}{|BC'|} \tag1$$ Consequently, $$\triangle CBC'\sim\triangle AHC'$$, so that $$\angle B=\angle AHC'=\angle A'HC$$. Thus, two angles of $$\triangle A'HC$$ match two angles of $$\triangle C'BC$$, so that the third angles must match, too. We conclude that $$AA'$$ is itself an altitude. $$\square$$