Below is described the construction of such an mixtilinear incircle contained in the angle A of triangle ABC.

As usual for such tangency problems, use an inversion. Here preferably to the circle A(2R), centered at A with radius 2R, R being the circumradius of ABC.

By this inversion the circumcircle of ABC maps to line DE, orthogonal to the diameter AA

The

The contact point A' of the mixtilinear with the circumcircle maps to A'', the contact point of the tritangent circle with side DE. Let M bet the intersection of AA' with BC.

In view of the similarity of triangles ABC and ADE, A''D/A''E=(s-c)/(s-b), s being the half-perimeter and {a,b,c} the side lengths opposite to {A,B,C} correspondingly.

Draw the parallel to DE from A, which is tangent to the circumcircle of ABC. Measuring the cross-ratio of the lines (AC,AB,AM,AG)=c through their intersections with lines DE and BC we find:

(i) On line DE, c = DA''/EA'' = (s-c)/(s-b).

(ii) On line BC c = (CM/BM):(CG/BG) = (CM/BM):(b

Equating we get CM/BM = (s-c)/(s-b)(b

(a/(b+c-a), b/(c+a-b), c/(a+b-c)),

the intersection point Q of AX with BC has CQ/BQ = (a+b-c)/(c+a-b)(b

Repeating almost verbatim the arguments of the preceding section we find the contact-point of the A-mixtilinear ex-circle as the intersection point A' of the line AX with the circumcircle. Here X is the triangle center X(55) coinciding with the inner similitude center of the incircle and circumcircle of ABC.

As in the previous section locate the position of M on BC and relate it to X(55). The A-mixtilinear ex-circle is circle d. Using the inversion with respect to the circle A(2R) this circle maps to the incircle of triangle ADE, line DE being the inverse of the circumcircle of ABC and similar to it.

Produced with EucliDraw© |