Triangle C = 2A

Construct a triangle ABC with angle C = 2A and side AB = 2AC. (Geometriagon #1031)

Let's complete the isosceles trapezoid ACBD, with ∠DAC = BCA.
AB is hence the angle bisector of DAC, hence DAB = BAC = ABD and triangle ABD is then isosceles, that is AD = BD.
As an isosceles trapezoid is a cyclic quadrilatéral, Ptolemy's theorem gives AB.CD = AC.BD + AD.BC
Naming x = AD = BD = BC, a = AC given and AB = CD = 2a : 4a² = ax + x² that is x² + ax - 4a² = 0, quadratic equation in x the solutions of which can be easily constructed, giving side BC of the searched triangle.

Construct AP = 2 AC.
On the perpendicular to AC from A, construct AO = AC/2.
Circle centered in O through P interesects this perpendicular in M and M'. (AM+AM')/2 = AO = a/2 and AM.AM' = AP² = 4a².
AM and AM' are then the solutions of the quadratic.
Circle centered in C with radius AM and circle centered in A through P intersect in the searched point B.
The choosen solution from the four possible being that with angle A acute, and of vourse the symmetric.
The solution with obtuse A results into C plus 180° (line angles modulo π) instead of inside angle to triangle.

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Some hyperbola...

The locus of points B with angle BCA = 2 BAC is an hyperbola.
The previous construction gives the intersection of this hyperbola with circle centered in A and radius 2 AC.
The proof comes directly from the previous trapezoid. The distance from B to perpendicular bisector of AC is then BC/2.
The locus of B is then hyperbola with focus C, directrix line the perpendicular bisector of AC and excentricity BC/BH = 2.
We deduce A is a vertex of this hyperbola, the other one being S with AS = 2/3 AC.
The center is I with AI = 1/3 AC, the other focus is F with AF = -1/3 AC. The asymptotes are at 60°.

 

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