## Carnot's (or Bottema's) Perpendicularity Theorem & Some Generalizations

Carnot's Perpendicularity Theorem
S is any point inside (or outside) triangle DHL and perpendiculars are dropped from S to the sides, and squares are constructed as shown.

Investigate & Conjecture
What do you notice about the two sums of the areas of the green squares and of the pink squares? Formulate a conjecture.

#### .sketch_canvas { border: medium solid lightgray; display: inline-block; } Carnot's Perpendicularity Theorem & Some Generalizations

Historical Note
I'd previously known this result as Bottema's Theorem (1938) & had called it thus, but apparently it is originally due to the French mathematician, Lazare Carnot (1753-1823). This result also appears as a problem in Challenging Problems in Geometry by Alfred Posamentier & Charles Salkind (1996), pp. 14; 85-86. It is an exercise in C V Durell's New Geometry for Schools (1939), p. 287, Q26. Earlier still, it is in J W Russell's Sequel to Elementary Geometry (1907), p. 34, Sect 6 - a worked example. The converse result is an exercise in a French textbook, Traite de Geometrie by E Rouche and C de Comberousse (1900) vol 1, p. 395, Q254.

Challenge
1) Can you explain why (prove) the theorem above is true? Hint: Connect S with the vertices and apply the theorem of Pythagoras to the six right triangles that are formed, group, and simplify.
2) Formulate the converse. Can you prove it?

Application & special cases
a) The concurrency of the perpendicular bisectors of a triangle, as well as of its altitudes, are special cases of the converse, which can be stated as follows: If DC2 + HG2 + LK2 = CH2 + GL2 + KD2, then the perpendiculars at C, G and K are concurrent at S.
b) The theorem of Pythagoras is a special case of Carnot's perpendicularity theorem. To visually see this in the figure above, drag triangle DHL until it is (approximately) a right triangle. Then drag the point S to place it at any one of the vertices other than the vertex of the (approximate) right angle.
c) The converse of the result can also be used to easily prove the concurrency of the Power Lines of a Triangle.
d) Different variations and applications of Carnot's Perpendicularity Theorem often appear in mathematics competitions and olympiads for high school learners. In fact, it was used as Q23 in Round 2 of the SA Mathematics Olympiad paper of 2023: From point P inside △ABC, perpendiculars are drawn to the sides meeting BC, CA and AB at points D, E and F, respectively. If BD = 8, DC = 14, CE = 13, AF =2√(30) and FB = 6, find AE.
(Though one can solve it by applying Pythagoras several times, it's straight forward to solve with the above theorem.)

Further Generalization
d) Can you generalize Carnot's Perpendicularity theorem further?
e) Click on the 'Link to Similar Figures' button in the above sketch. Dynamic similar rectangles on the sides are shown, but just like the theorem of Pythagoras itself, the result generalizes to any similar figures, e.g. half circles or regular pentagons, etc. on the sides. Can you prove it in general? What about the converse? Is it true? Can you prove or refute the converse?
f) Click on the 'Link to Polygon' button in the above sketch. A quadrilateral is shown but the result generalizes similarly to any polygon. Can you prove it in general? What about the converse? Is it true? Can you prove or refute the converse? What about similar figures on the sides of any polygon?