Prove that Quadrilateral ABCD is a Rhombus
show all the work

Say that the shape is a parallelogram with equal length sides;

AB = CD

ABD = CDB as the bisecting lines (you have drawn) prove ABD : CDB are both isosceles triangles.

Therefore ABD = ADB

Draw the shape's diagonals as each others' perpendicular bisectors;  

Draw the shape's diagonals bisect both pairs of opposite angles.

If two non congruent angles were given in example 70 degree and 110 degree then say M ∴ = (1/2 )(M∴BAD) = 35  =  (1/2 )(M∴BEF) = 35

and do the same for  1/2(M∴ABC) = 55 = 1/2 (M∴CDA) = 55

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