Orthic

1. Orthic triangle

Given triangle ABC, the orthic triangle DEF is formed by the feet of the altitudes of ABC. The orthocenter is the common point H of these altitudes. The following facts are well known:
[1] The circle k_A with diameter AH passes through E, F. Analogously k_B, k_C pass through {F,D} and {D,E} correspondingly.
[2] Triangles AFE, BDF, CED are similar to ABC.
[3] The altitudes of ABC are bisectors of DEF.
[4] The angles of DEF are related to those of ABC: angle(D) = pi - 2*angle(A). Analogously the other pairs of opposite angles.
[5] |FE| = |BC|*cos(A). Thus, cos(A) is the similarity ratio of AFE to ABC. Analogous relations for the other side-pairs of DEF and ABC.
[6] |BC| = |AH|*tan(A). Analogous relations etc..
[7] The tangents t_A, t_B, t_C of the circumcircle (c) of ABC are parallel to corresponding sides of DEF. Hence the radius OA is orthogonal to side EF of the orthic, and analogous relations for the other sides of the orthic.
[8] The symmetrics A'', B'', C'' of H with respect to the sides of ABC are on the circumcircle (c) of ABC.
[9] AH*HD = BH*HE = CH*HF.
[10] The circumcircle (e) of DEF is the Euler circle of ABC. Later is homothetic to (c) with respect to H at ratio 1/2.

[1] Since points {E,F} view AH under a right angle, they are on circle k_A. Analogous proofs for {k_B, k_C}.
[2] Since points {E,F} view also BC under a right angle they are on the circle with diameter the side BC. It follows that angle(AFE) = angle(C), proving that AFE and ABC are similar triangles. Analogous proofs for triangles BFD and EDC.
[3] By [1] angle(FDH)=angle(FBH) and angle(HDE)=angle(HCE). But angle(FDH)=angle(HCE) since both together with angle(A) make a right angle.
[4] Follows by [1] and [3] since angle(FDE)=pi-angle(BDF)-angle(EDC) etc..
[5] Draw a parallel FF' to AC, F' being on the circle with diameter BC, and measure EF/BC = CF'/BC = cos(A), since angle(BCF') = angle(BFF') = angle(A).
[6] AH/BC = F'H'/BC, H' being the projection of F' on BC etc..
[7] angle(t_A,BC) = angle(C) = angle(AFE), because of [1] etc..
[8] Assume A'' is the other intersection point of AD with the circumcircle and show that BHC and BA''C are equal triangles, having common the side BC and equal angles at B and C.
[9] Measure the power of point H with respect to circles with diameters respectively {AB,BC,CA}. This common product is relevant for the discussion of autopolar triangles and the orthocentric anti-inversion discussed in OrthocentricAntiInversion.html .
[10] Follows from the discussion on the Euler circle (see Euler.html ) .

2. Inside out view

Starting form DEF we can obtain ABC by drawing the bisectors of DEF. H is then the incenter of DEF and {A,B,C} are the centers of the excircles of DEF. The part AA₁ of the circumradius is the radius of the excircle centered at A. Analogous properties hold for the radii of the other excircles centered at B and C. This aspect of the previous figure is discussed in Bisector1.html .
[1] Since points {E,F} view AH under a right angle, they are on circle k_A. Analogous proofs for {k_B, k_C}.
[2] Since points {E,F} view also BC under a right angle they are on the circle with diameter the side BC. It follows that angle(AFE) = angle(C), proving that AFE and ABC are similar triangles. Analogous proofs for triangles BFD and EDC.
[3] By [1] angle(FDH)=angle(FBH) and angle(HDE)=angle(HCE). But angle(FDH)=angle(HCE) since both together with angle(A) make a right angle.
[4] Follows by [1] and [3] since angle(FDE)=pi-angle(BDF)-angle(EDC) etc..
[5] Draw a parallel FF' to AC, F' being on the circle with diameter BC, and measure EF/BC = CF'/BC = cos(A), since angle(BCF') = angle(BFF') = angle(A).
[6] AH/BC = F'H'/BC, H' being the projection of F' on BC etc..
[7] angle(t_A,BC) = angle(C) = angle(AFE), because of [1] etc..
[8] Assume A'' is the other intersection point of AD with the circumcircle and show that BHC and BA''C are equal triangles, having common the side BC and equal angles at B and C. [9] Measure the power of point H with respect to circles with diameters respectively {AB,BC,CA}. This common product is relevant for the discussion of autopolar triangles and the orthocentric anti-inversion discussed in OrthocentricAntiInversion.html . [10] Follows from the discussion on the Euler circle (see Euler.html ) .
Remark
The orthic triangle is related to other well known problems, such as:
a) The Fagnano problem by which we seek the triangle with least perimeter inscribed in an accute-angled triangle.
b) The Billiard ball problem by which we seek the shortest closed path of a billiard ball reflecting on the sides of ABC.
c) The Three reflexions composition by which we seek to describe the composition of reflexions on three given lines in general position (forming the sides of a triangle ABC).
All these problems are discussed in the references given below.

1. Orthic triangle

2. Inside out view

See Also