angles in rhombus when an equilateral triangle is inscribed in it [closed]

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When one inscribs an equilateral triangle in a rhombus, all the corners are multiples of 30 degrees. I can see this, but I can't proof it.

Question: How can I proof that the angle ADC is 120 degrees?

And how do I claim that the triangle SHB (S being the intersection of the diagonals) is similar with triangle FHC? I can see it and I think there is an easy way with parallel lines (why is EF || AC) and bisectors (the diagonals of a rhombus) but I just can not put my finger on it...

Any help and/or hints are really appreciated.

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2 Answers

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The angle aren't multiples of 30 degree. It can be any value you like. If $D$ is moved upwards then $A$ and $C$ contract inwards (and move upwards as well) and vice versa.

Some examples:

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It is incorrect. Rhombuses of varying aspect ratio are possible to be circumscribed through $DAC$ with a variable height $BD$.

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