Which triangles must be equilateral? $\triangle ABE$ and $\triangle CDE$? $\triangle ABC$ and $\triangle ABD$? or ?

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parmenides51 wrote: An isosceles trapezoid is divided by it's diagonals into two equilateral triangles. Determine the angles of the trapezoid. the correct translation (according to the given solution) should be: An isosceles trapezoid is divided by each diagonal into two isosceles triangles. Determine the angles of the trapezoid.

original wording

@above, thanks again

Let $A(0,0),B(b,0),C(b-a,c),D(a,c)$. In $\triangle ABC\ :\ AB=AC \Rightarrow (b-a)^{2}+c^{2}=b^{2}$. In $\triangle ACD\ :\ AD=CD \Rightarrow a^{2}+c^{2}=(b-2a)^{2}$. Solving $a=\frac{(3-\sqrt{5})b}{4}$ and $c=\frac{b\sqrt{10-2\sqrt{5}}}{4}$. Then $\tan \angle BAD = \sqrt{5+2\sqrt{5}}$ and $\angle BAD=72^{\circ}$.

Triangles $ABC$ and $ACD$, $ABD$ and $CBD$ all must be isosceles, so it seems $ABCD$ to be a square! I however might be wrong, maybe there are other few possibilities!

$8$ isosceles triangles.

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