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

stop cheating

step-by-step explanation:

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.