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.

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.

Related Links
Power Lines Special Case: Altitudes of a Triangle
Water Supply II: Three Towns (Rethinking Proof activity)
The Center of Gravity of a Triangle (Rethinking Proof activity)
Triangle Altitudes (Rethinking Proof activity)
The Fermat-Torricelli Point (Rethinking Proof activity)
Airport Problem (Rethinking Proof activity)
Napoleon (Rethinking Proof activity)
Miquel (Rethinking Proof activity)
Kosnita's Theorem
Fermat-Torricelli Point Generalization (aka Jacobi's theorem) plus Further Generalizations
Dual to Kosnita (De Villiers points of a triangle)
Another concurrency related to the Fermat point of a triangle plus related results
Experimentally Finding the Medians and Centroid of a Triangle
Experimentally Finding the Centroid of a Triangle with Different Weights at the Vertices (Ceva's theorem)
Point Mass Centroid (centre of gravity or balancing point) of Quadrilateral
Bride's Chair Concurrency
Triangle Centroids of a Hexagon form a Parallelo-Hexagon: A generalization of Varignon's Theorem
Nine-point centre & Maltitudes of Cyclic Quadrilateral
A side trisection triangle concurrency
A 1999 British Mathematics Olympiad Problem and its dual
Concurrency and Euler line locus result
Haag Hexagon - Extra Properties
Concurrency, collinearity and other properties of a particular hexagon
Carnot's Perpendicularity Theorem & Some Generalizations
Generalizing the concepts of perpendicular bisectors, angle bisectors, medians and altitudes of a triangle to 3D
Anghel's Hexagon Concurrency theorem
Some Circle Concurrency Theorems
Three Overlapping Circles (Haruki's Theorem)
Parallel-Hexagon Concurrency Theorem
Toshio Seimiya Theorem: A Hexagon Concurrency result
Van Aubel's Theorem and some Generalizations (See concurrency in Similar Rectangles on sides)
The quasi-circumcentre and quasi-incentre of a quadrilateral

External Links
Concurrent lines (Wikipedia)
SA Mathematics Olympiad Questions and worked solutions for past South African Mathematics Olympiad papers can be found at this link.
(Note, however, that prospective users will need to register and log in to be able to view past papers and solutions.)

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