[1] The circle k

[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

[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

[1] Since points {E,F} view AH under a right angle, they are on circle k

[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

[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

[10] Follows from the discussion on the Euler circle (see Euler.html ) .

[1] Since points {E,F} view AH under a right angle, they are on circle k

[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

[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

The orthic triangle is related to other well known problems, such as:

a) The

b) The

c) The

All these problems are discussed in the references given below.

BilliardClosed.html

Euler.html

Fagnano.html

OrthocentricAntiInversion.html

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