Vtyjkyuk

I have an isosceles trapezoid $ABCD$.

In my development, I have obtained that the parallel sides are equal. But that is impossible, because if that happened, it would be a parallelogram.

So, what is the error in my development? Thanks in advance.

Basically, $DC$ can't be equal to $AB$, so what is my mistake?

Are you given those angles are equal?

If so then the trapezoid IS a parallelogram that is not drawn to scale. (Who told you that the trapezoid wasn't a parallelogram?)

The angles marked $beta$ must be equal because $overlineDC$ is parallel to $overlineAB$. But the angles marked $alpha$ are equal if and only if $overlineAD$ is parallel to $overlineBC$.

Which if this is drawn to scale they are not; but which otherwise you have no reason to rule out.

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Ah. I see. $angle DCA = beta$ but $angle ACB ne alpha$.

Because it is a isoceles trapezoid the base angles are congruent. Then base angles are $angle ABC$ and $angle DAB$ so $angle ABC = angle DAB =alpha + beta$.

$angle DBC$ is NOT a base angle. And $angle DBC ne angle DAB$ (even though $angle DCA = beta$.)

Peter's right that you don't have similar triangles, but here's another way to check. A sanity check, if you will.

A defining characteristic of similar triangles is that their angles have the same measures. It's easy to see that $triangle ACD$ has an obtuse angle, while $triangle CAB$ does not. Therefore, there's no way the two can be similar.

The angle $alpha$ does not occur twice because $AD$ and $BC$ are not parallel.

Hence you do not have similar triangles.