The following proof is based on a plane quadrilateral. If it is a space, ask again.

Connecting BD, we can get from sine theorem:

In △ABD，BD/sin∠A = AB/sin∠ADB；

In △CBD, BD/sin∠C=CD/sin∠CBD.

And ∠ A = ∠ C because AB=CD.

So ∠ADB=∠CBD or ∠ ADB = 180-∠ CBD,

ADB can be proved when ∠ADB=∠CBD. From the inner angle of the triangle, it can be concluded that the other two angles are equal ∠ABD=∠CDB, which can prove ABD. This is a parallelogram.

When ∠ ADB = 180-∠ CBD, it is not established.

The diagram of the counterexample is like this.

Downstairs is one of them, AB=CD=√3, BC=2, AD= 1, ∠ A = ∠ C = 30.

The following figure shows that AB=CD, ∠A=∠C in general. What should I do with this picture?

Draw an obtuse triangle ABD first, then make a circle with D as the center and AB length as the radius, and finally make ∠ DBC = 180-∠ ADB with point B as the vertex, where BC intersects the circle at point C.

The quadrilateral ABCD satisfies AB=CD, ∠ A = ∠ C.

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