Triangle \(ABC\) is drawn so that the angle bisector of \(\angle{BAC}\) meets \(\overline{BC}\) at \(D\) and so that triangle \(ABD\) is an isosceles triangle with \(AB=AD\). Line segment \(\overline{AD}\) is extended past \(D\) to \(E\) so that triangle \(CDE\) is isosceles with \(CD = CE\) and \(\angle DBE = \angle DAB\). Show that triangle \(AEC\) is isosceles.
There are 4 isosceles triangles: ABD, CDE, BCE, and AEC.
Δ ABD => 45° + 67.5° + 67.5° = 180°
Δ CDE => 45° + 67.5° + 67.5° = 180°
Δ BCE => 45° + 90° + 45° = 180°
Δ AEC => 45° + 67.5° + 67.5° = 180°
Triangle AEC is an isosceles triangle, because 2 of its 3 angles are equal, and as a result of that, 2 of its sides are equal as well.