Given the triangle t = ABC, consider the intersection points G, H of the medial lines FO, DO with sides AC, AB respectively.

1) q = BCGH is a cyclic quadrilateral (angle(HCG) = angle(HBG) = angle(A)).

2) The circumcircle c of q passes through the circumcenter O of the triangle t.

3) The circumcircle d of AFD and c intersect at O and E, such that A, E, I are on a line. I being the intersection of the medial of BC and c, diametral to O (orthogonality of OE to line AE).

4) angle(A) = angle(BGI) = angle(BHI) = angle(BEI) = angle(IEC) = angle(IGC) and p = AHIG is a parallelogram.

5) HG is bisected by AI at M, hence AI is the symmedian w.r. to A (see Antiparallels.html ).

6) angle(FAE) = angle(FOE) = angle(GCE) and since, because of the symmedian AI, angle(JAC) = angle(FAE) the triangle ANC is isosceles and N is on the medial of AC.

7) Similarly APB is isosceles and P is on the medial line of AB. Hence E is also the intersection point of the sides BP and CN of the isosceli ABP, ACN, defined through the median AJ.

8) angle(IBC) = angle(ICB) = angle(A) leads to an easy construction of the symmedian AI.

9) angle(BAE) = angle(ECA), angle(ABE) = angle(EAC) shows that ABE and CAE are similar triangles.

For an application of these remarks in a case of determination of the focus of a parabola, look at the file ParabolaSkew.html .

Antiparallels.html

Cosymmedian.html

ParabolaChords.html

ParabolaProperty.html

ParabolaSkew.html

Symmedian.html

Symmedian_0.html

Symmedian_Vecten.html

SymmedianProperty.html

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