prove that quadrangle is isosceles trapezoid How to prove that quadrangle $ABCD$ is a isosceles trapezoid? 
where $AB$ is parallel to $CD$

 A: $AB$ is parallel to $CD$ and then, $\angle BCD=\angle BAD=\angle ADC$. thus, $BD=AC$
A: Well, since $AB$ is parallel to $CD,$ then you already know it is a trapezoid. Note that the segments drawn to the center $O$ from the vertices of the trapezoid split it into isosceles triangles. If you can show that $\angle AOC$ and $\angle BOD$ have the same measure, it will follow that $AC$ and $BD$ are of the same length, as desired.
A: You are given two facts:
    1. A qualrilateral ABCD is inscribed in a circle.
    2. AB is parallel to CD

Because of the parallel lines, qualrilateral ABCD is, by definition, a trapezoid.
To prove that trapezoid ABCD is isosceles, you need to show that the non-parallel sides BD and AC have equal lengths.  This can be accomplished as follows.  Erase the lines that go to the circle's center. Draw a single line from point A to point C.  Finish the proof by using these three theorems:
    1. If parallel lines are cut by a transversal, 
       then the alternate interior angles measure the same.

    2. In a circle, the arc subtended by an inscribed angle has
       twice the measure of the inscribed angle.

    3, In a circle, if two minor arcs are equal in measure, 
       then their corresponding chords are equal in length.

A: There are two given facts:inscribed quadrilateral and AB is parallel to DC
From 1st given fact,
We can know that the sum of opposite angles of inscribed quadrilateral is 180.
angle DAB +angle BCD = 180
From 2nd,
Since AB is parallel to DC,
angle ABC + angle BCD = 180
So, we get angle DAB + angle BCD = angle ABC + angle BCD
        Therefore,                angle DAB = angle ABC
