# Orthocenter: The "Bad Boy" of Distinguished Points in a Triangle

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. The proof known to me (see eg here) involves the construction of an auxiliary triangle $$A'B'C'$$ that contains $$ABC$$ in a suitable manner and with the property that the altitudes of $$ABC$$ correspond to bisectors of $$A'B'C'$$. And that the bisectors of a triangle intersect at a point can be seen "immediately": For example, look at the intersection of two bisectors through $$A'$$ and $$B'$$, then by construction, this must lie on the remaining bisector through $$C'$$, since it has the same distance from $$A'$$ and $$B'$$.

What I'm wondering, however, is whether there isn't a synthetic "direct" - and as well "coordinate-free" and without analytic methods - proof that the altitudes intersect at exactly one point, ie which does not rely on "artificial" auxiliary constructions, but rather more or less "immediately" on elementary tools like congruence & similar triangle theorems, elementary angle relations (“angle chasing”) ?

The point is that seemingly the proof of existence orthocenter appears to be much more "unnatural" then for other "distinguished" points of triangles, like incenter, circuscenter and centroid.

The proofs for all three are "immediate" based on same argument, on draws two lines and sees immediately "by definition" that the third line intersects the other two exactly in same point where the two intersect. Ok for centroid one has to invoke the Intercept theorem, but that's basic.

On the other hand the orthocenter seemingly - at least I don't know a "straight forward" proof - uses some "external" or auxilary constructions or external more or less "non trivial" results like Ceva theorem.

So this rases the philosophical question why in the proofs for existence of orthocenter is seemingly much more involved then in other distinguished points of a triangle?

• One can let the altitudes from vertices $B$ and $C$ intersect at point $H$, and then prove that $AH$ is perpendicular to $BC$. This proves that the three altitudes are concurrent. Would this be a "straightforward enough" proof that you are looking for? Commented Jul 20 at 20:42
• @Euclid: Yes, but is it "immediate" that $AH$ is perpendicular to $BC$? Commented Jul 20 at 20:45
• @Euklid: Ie, is it clear that the line $AH$ intersects $BC$ orthogonally? That's what I meant by "angle chasing approach" in suggested: I tried to see that the angles between $AH$ and $BC$ are right angles, but failed to prove it directly by such naive "angle chasing" Commented Jul 20 at 20:50
• @user267839 It is not immediate by "naive angle chasing", but it is immediate by angle chasing through cyclic quadrilaterals, if you consider identifying cyclic quadrilaterals through equal angles as "immediate" Commented Jul 20 at 20:52
• There are hundreds of other remarkable points (Nagel point, symmedian point, nine-point center, Feuerbach point, Kosnita point, etc.), whose proofs are generally even deeper than that for the orthocenter. It's the three easy one that are outliers! Commented Jul 20 at 21:16

I don't know if you will like this proof, which should IMHO be considered as a coordinate-free one ...

For all point $$M$$ in the plane :

$$\begin{eqnarray}\overrightarrow{AM}.\overrightarrow{BC}+\overrightarrow{BM}.\overrightarrow{CA}+\overrightarrow{CM}.\overrightarrow{AB}&=&\overrightarrow{AM}.\overrightarrow{BC}+\left(\overrightarrow{AM}-\overrightarrow{AB}\right).\overrightarrow{CA}+\left(\overrightarrow{AM}-\overrightarrow{AC}\right).\overrightarrow{AB}\\ & = &-\overrightarrow{AB}.\overrightarrow{CA}-\overrightarrow{AC}.\overrightarrow{AB}\\ &=&0\end{eqnarray}$$

Hence, if $$M$$ lies on the altitude from $$A$$ and also on the altitude from $$B$$, then $$\overrightarrow{AM}.\overrightarrow{BC}=0$$ and $$\overrightarrow{BM}.\overrightarrow{CA}=0$$, so the previous identity gives : $$\overrightarrow{CM}.\overrightarrow{AB}=0$$. This proves that $$M$$ lies on the altitude from $$C$$ too.

It should be added that the altitudes from $$A$$ and from $$B$$ meet at least at one point, since $$ABC$$ is a "true" triangle.

Let $$BD$$ and $$CE$$ be the altitudes that meet at $$H$$. We'll show that $$AH$$ is an altitude too, which will prove that the three altitudes in a triangle are concurrent.

Let $$AH$$ intersect $$BC$$ at $$K$$.

Notice that due to the right angles at $$D$$ and $$E$$, the circle with diameter $$BC$$ passes through $$D$$ and $$E$$. Likewise, the circle with diameter $$AH$$ passes through $$D$$ and $$E$$. In other words, the quadrilaterals $$BCDE$$ and $$ADHE$$ are cyclic.

\begin{align} \angle HBK + \angle BHK &= \angle DBC + \angle AHD \\ &= \angle DEC + \angle AED \\ &= \angle AEC \\ &= 90^\circ. \end{align}

Therefore, $$\angle BKH = 180^\circ - 90^\circ = 90^\circ$$ too, i.e., $$HK \perp BK$$, i.e., $$AH \perp BC$$, as desired. $$\blacksquare$$

Let the feet from A,B,C to their opposite sides be D,E,F. Note that by Thales' theorem that quadrilaterals BCEF, ABDE, ACDF are cyclic quadrilaterals. Then by the Radical Axis theorem, the radical axes of all of these circles must concur, so AD, BE, CF must concur.

Let altitudes $$BB'$$ and $$CC'$$ meet at $$H$$, and let $$A'$$ be the foot of the perpendicular from $$H$$ (but not necessarily from $$A$$) to $$BC$$. Define $$a:=|BC|$$, $$b:=|AC|$$, $$a':=|A'C|$$, $$b':=|B'C|$$.

Easy angle-chasing shows that $$\triangle HA'C\sim\triangle BC'C$$ and $$\triangle HB'C\sim\triangle AC'C$$ (via Angle-Angle-Angle). Therefore, $$aa' \underbrace{\qquad=\qquad}_{\triangle HA'C\;\sim\;\triangle BC'C} |CH||CC'| \underbrace{\qquad=\qquad}_{\triangle HB'C\;\sim\;\triangle AC'C} bb'$$ This implies $$\triangle CAA'\sim\triangle CBB'$$ (via proportionality); in particular, $$\triangle CAA'$$ has a right angle at $$A'$$. We conclude that the perpendicular from $$H$$ to $$BC$$ is indeed necessarily also the perpendicular from $$A$$, so that $$H$$ lies on all three altitudes. $$\square$$

Note. One can re-work the argument above to initially take $$A'$$ as the foot of the perpendicular from $$A$$, showing that it's also the foot of the perpendicular from $$H$$.

This underscores the fact that $$A$$ is itself the orthocenter of $$\triangle HBC$$. (Likewise, $$B$$ and $$C$$ are the orthocenters of $$\triangle AHC$$ and $$\triangle ABH$$, so that all four points comprise an orthocentric system.)

• That kind of argument I was exactly looking for, thanks! One nitpick: Do you know if it is possible to modify the argument appropriately to reasoning "reversely" using similar methods? Namely as above we draw altitudes $BB'$ and $CC'$ to be "given as input" intersecting in $H$ as above and consider instead as second input the line $AH$ which intersects $BC$ in internal complementary angles at $A'$ ( ie "inside" $ABC$) $\alpha_1, \alpha_2$ with $\alpha_1+\alpha_2= 180$ by construction, but a priori we not know that $\alpha_1, \alpha_2=90$ ( ...thats what we want to show) Commented Jul 21 at 9:56
• Could you modify your argument slightly to approach it from the "other side"? Essentially this would differ from your presented approach above in only one point: You assumed as "input" the line $A'H$ to be perpendicular/ foot to $BC$ and show that it actually goes through $A$. But the question is if this argument cannot be slightly modified to instead to start with assumption/ input $AH$ to be given with intersection with $BC$ at $A'$ a priori non perpendicular and then deduce by similar reasonings the perpendicularity? Commented Jul 21 at 10:02
• (I hoped that once having drawn $ABC$ together with $BB'$, $CC'$ and $AH$ the perpendicularity of $AH$ and $BC$ can be similarilyas in your reasonings above more less read up from proportionality considerations; but it appears to me -so far I see -that it becomes unlikely harder if we slightly swap the assumption by instead to work with $HA'$ given as foot to the line $AH$). Is there any pathology going on in background why such slight changing of "what we assume" changes "dramatically" the argumentation. Note that similarity of $B'CH$ and $C'HB$ is immediate. Can we proceed somehow from this? Commented Jul 21 at 10:23
• ...As well congruence of $AC'C$ and $AB'B$ is obvious. But from this I not know how to proceed to draw conclusion that $\alpha_1 =\alpha_2=90$. ( at least I have to invoke here method of cyclic quadrilateral, but it question is if it can be also done without; so just by congruent triangles reasonings in similar way as in your reasonings above?) Commented Jul 21 at 10:36