As shown in the figure, in trapezoidal ABCD, ab ‖ CD, ∠ a = 51 ° and ∠ B = 78 ° prove: CD + BC = ab

As shown in the figure, in trapezoidal ABCD, ab ‖ CD, ∠ a = 51 ° and ∠ B = 78 ° prove: CD + BC = ab


It is proved that a parallel line AE parallel to ad is drawn through point C, and ab intersects point e. the AECD is a parallelogram, the AECD is a parallelogram, the AECD is a parallelogram, and the corresponding sides of the parallelogram are equal. In △ CEB, the AECD is a parallelogram



In quadrilateral ABCD, if vector AB = vector BC and absolute value AB = absolute value ad, what shape is quadrilateral ABCD
Why? It's better to give a reason
I don't think it should be a parallelogram? How did it change into a diamond?
(a quadrilateral with parallel and equal sides is a parallelogram)


First of all, we should know that diamond is a kind of parallelogram
This is a parallelogram, but "a group of parallelograms with equal adjacent sides is a diamond," so this is a diamond