Question 981938
Let me start this with AB = AD = 10 and BC = CD = 5

Prove AB = AD > BC = CD

If AB = AD and BC = CD then ACD = ABC => A(CD) = (AB)C => A(C)D = A(BC) => 

A(5) = (10)C => (10)C = (5)A => C = 5/10A  => C = 1/2A thus 2C = A

(2C)B  = (2C)D  = 2(CB) = 2(CD) =>  Since A = 2C and  BAD = B(2C)D =>  

BC(2D) => BAD =>(5)(2)D = 10D => BAD => B(10) => BAD thus 10D = 10B = BAD and 

D = B => BA  = AD  =  10 then B = 1 and D = 1 thus BAD = 10 because (1)(10) = 10 

End line

Shape of a kite, boomerang, or a crossed quadrilateral

Not sure whether I did it correctly so let me know if this is what it should have been or not.

ED