Draw from F a parallel to CD, defining point I on CE, such that DF/FE = CI/IE = k. Define also the points G and H on AC and BC respectively, such that AG/GC = CH/HB = k. Given k, G and H are well defined points on the sides of t and triangle s = IFH is equal to CGH.

Notice that extending the other sides of the isoscelli: BE, CD, etc. the triangle t* = A*B*C* defined through their intersections is similar to t and shares with it the same first Brocard point M. M being the intersection point of the three circles {ABB*}, {BCC*} and {CAA*}. These circles being tangent respectively to BC, CA and AB. See the file Brocard.html on Brocard points.

Produced with EucliDraw© |