Generalization of Pythagorean Theorem with Orthocentre

We all know Pythagorean Theorem: the legs a, b and a hypotenuse cc of any right triangle satisfy an equation: a2+b2=c2a^2 + b^2 = c^2

The commonly known generalization of that theorem is Law of Cosines:

c2=a2+b22a×b×cos(C) c^2=a^2 + b^2 - 2 a \times b \times cos(C)

There is another generalization and it involves use of Orthocentre. Let HH be an Orthocentre of ABC\triangle ABC, let D,E,FD,E,F denote feet of altitudes from A,B,CA,B,C respectively.

cos(C)=CEa=CDbcos(C) = \dfrac{CE}{a} = \dfrac{CD}{b}

When we use that as a substitution in the Law of Cosines we obtain: c2=a(aCD)+b(bCE)=a×BD+b×AEc^2 = a(a-CD) + b(b-CE) = a \times BD + b \times AE or AB2=AC×AE+BC×BDAB^2 = AC \times AE+ BC \times BD

Since ADE=90,BEC=90\angle ADE = 90, \angle BEC = 90 we conclude that quadrilateral CEHDCEHD is cyclic. Using power of point we can state that: AE×AC=AH×ADAE \times AC = AH \times AD and BD×BC=BH×BEBD \times BC = BH \times BE

Susbstituting it in the previous equation we get

AB2=AH×AD+BH×BEAB^2 = AH \times AD+ BH \times BE

In a special case when C=90\angle C = 90 then H=C=D=EH=C=D=E and it becomes regular Pythagorean Theorem.

What is interesting is that if we create a right triangle on the other side of ABAB, with legs AP=AH×AD,BP=BH×BEAP=\sqrt{ AH \times AD}, BP= \sqrt{ BH \times BE} and hypotenuse ABAB, and repeat for sides BC,ACBC, AC we will get a net of a orthogonal tetrahedron with apex PP.

Orthogonal projection of apex PP onto ABC\triangle ABC is orthocentre HH. Projection of APAP is AHAH etc.

It is commonly known that AH×HD=BH×HE=CH×HFAH \times HD= BH \times HE = CH \times HF.

It can be easily shown when exploring tetrahedron ABCPABCP that this constant is HP2HP^2.

HP2=AH×HD=BH×HE=CH×HF HP^2 = AH \times HD= BH \times HE = CH \times HF


I currently do not have time to fill out all the details but I trust anybody who knows basics of geometry can fill in the blanks. I have come into realization of these facts which amuse me some years ago when solving brilliant Trirectangular Corner Locus and eventually I made an effort to write this as imperfect as it is. I thought it is better than not doing it at all. Writing, even notes or solutions can always be perfected, but fear of imperfection often means not doing it at all.

Note by Maria Kozlowska
2 years, 11 months ago

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Some notes:

  1. At the point of substitution of the results of cos(C)\cos(C) into the Law of Cosines, you will need to expand on how you arrived at the answer. You cannot expect people to get exactly what you are talking about, especially when you jumped at least 3-4 steps.

  2. I think what you are trying to get at is the Power of a point theorem; you will need to explain clearly what this is. Again, you can't just expect people to know what you are talking about if you're not willing to define your terms properly; a link would generally suffice.

  3. Nice observation with regards to the net of an orthocentric tetrahedron! We can see this to be true because the sum of squares of opposite edges of such orthocentric tetrahedron would be equal, by the eighth equation (you may want to number your equations for cleanliness, but I normally don't do this).

However, I don't see how this is a generalisation of Pythagoras' theorem using orthocentres. All you have done so far is use the orthocentre of a triangle to generate a net for an orthocentric tetrahedron...

P.S. If it takes me more than 10 minutes to try to understand this note, then I generally consider there to be something wrong with it. In general, if you lay out your steps properly, concisely and in detail, you will find that it would save the reader a heap of time to understand what you are writing.

A Former Brilliant Member - 2 years, 11 months ago

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May I ask whether you have written any papers on it? I am writing a paper on orthocentric tetrahedra and would really like to look into this aspect of it.

A Former Brilliant Member - 2 years, 10 months ago

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No, not yet,

Maria Kozlowska - 2 years, 10 months ago

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