## On the triangle's bisectors

Consider the incenter E of a triangle ABC. Let D be the point of contact of the incircle with the side BC. Then the line KL, which is orthogonal to AD at D, cuts the bisectors at B, C at points K, L, respectively, symmetric with respect to D.

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 .