## Power Lines of a Triangle

If triangles DBA, ECB and FAC are constructed outwardly on the sides of any triangle ABC so that DA = FA, DB = EB and EC = FC, then the perpendicular from D to AB, the perpendicular from E to BC and the perpendicular from F to AC are concurrent.

#### .sketch_canvas { border: medium solid lightgray; display: inline-block; } Power Lines of a Triangle

Special cases
a) Note that this result can be viewed as a generalization of the concurrency of the perpendicular bisectors of a triangle. To see this, click on the 'Show Perpendicular Bisectors' button & drag the configuration until all three points D, E and F respectively lie on the perpendicular bisectors of ABC.
b) This result is also a generalization of the concurrency of the altitudes of a triangle & hence provides an immediate proof. See for example, Special case - altitudes of triangle where applying the above theorem to the triangle DEF formed by the midpoints D, E and F of the sides of triangle ABC produces the desired result.

Challenge: Can you explain why (prove) the general theorem above is true? Can you explain (prove) it in different ways?

Application: This result, together with Ceva's theorem, can be used to develop a straight forward proof of Haruki's theorem.

1) For a proof using the concept of the power of a point, read the pp. 198-199 excerpt from my Some Adventures in Euclidean Geometry book at power lines proof.

2) For a simple proof using 3D geometry, read p. 7 of my joint Learning & Teaching Mathematics paper in 2008 with Mary Garner from KSU at Problemsolving and proving via generalization.

Carnot's perpendicularity theorem
The powerlines result can also be seen as a special case of the 'perpendicularity' theorem of the French mathematician, Lazare Carnot (1753-1823). With reference to the above figure, Carnot's perpendicularity theorem states that if T is any point and the feet of the perpendiculars from T to the sides AB, BC, CA are respectively labelled as G, H and I, then AG2 + BH2 + CI2 = GC2 + HC2 + IA2. The converse is also valid; hence by showing that AG2 + BH2 + CI2 = GB2 + HC2 + IA2, from the given DA = FA, DB = EB and EC = FC, the concurrency of the 'power lines' above follows easily from this theorem.
(Ironically, I've known Carnot's theorem as Bottema's theorem (1938), and have had a webpage about it, and some generalizations, on my Student Explorations page since 2009. However, I've now updated this Bottema webpage to WebSketchpad, and more historically correctly, named it Carnot's perpendicularity theorem.)

Challenge: Can you prove Carnot's perpendicularity theorem? Hint: Connect T with the vertices and apply the theorem of Pythagoras to the six right triangles that are formed, group, and simplify.

Michael de Villiers, created 14 July 2008; modified/adapted to WebSketchpad 30 August 2021; modified 1/3 September 2021.