The following remarks lead to the proof.

[1] The circle through JDB cuts line GD at F. BFEJ being cyclic, the angle at F is a right one.

[2] EF passes through A, thus F is on the bisector of A. In fact, angle(BEF)=angle(FDC)=(pi-C)/2. angle(BEA)=pi-(A+B)/2. Hence the two angles at E add to pi.

[3] The orthogonal triangles AGE and AFB are similar. Thus, considering a triangle AHI, similar to AGE and having its vertex H moving on DG, the other vertex I moves on line BE, i.e. the bisector of B (look at Moving_Similar_Polygons.html for the background).

[4] When H coincides with D, then I becomes coincident with K and triangles ADK, AFB, AGE are similar.

[5] Let L be the symmetric of K, with respect to D. Then triangles AKJ and ALG are equal. They have angle(KAJ) = angle(LAG) and equal sides AK = AL, AJ = AG. The equality of angles holds because angle(KAL) is equal to angle A of the triangle ABC.

[6] Since GL = JK and DL = KD = KJ , we have GL = DL and L is on the medial line of GD, which is the bisector of angle C.

For an application of this result to the symmedians of a triangle, look at SymmedianProperty.html .

Bisector0.html

Bisector1.html

BisectorRectangle.html

Euler.html

Moving_Similar_Polygons.html

SymmedianProperty.html

TriangleBisectorConstruction.html

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