[alogo] Angles on sides

Consider triangle t = ABC and an arbitrary angle v = JKL. Construct angles equal to v on the sides of ABC: ang(CBE) = ang(ACD) = v, so that triangles CBE and ACD are isoscelli. Then the locus of the point F, dividing DE in a fixed ratio k is a circle.

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]
[2_0] [2_1] [2_2] [2_3]

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.

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