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 *AG ^{2} + BH^{2} + CI^{2}* =

(Ironically, I've known Carnot's theorem as Bottema's theorem (1938), and have had a webpage about it, and some generalizations, on my

**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.

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