Quadrilateral ABCD is inscribed in a circle.

This is a cyclic quadrilateral.....the opposite angles are supplementary  (add to 180 degrees)

So

(x + 15)  + ( 4x +  5  ) =   180     simplify

5x  +  20   =  180     subtract 20 from both sides

5x  =  160       divide both sides by 5

x  = 32

So

A  =  4(32) + 5   =   133°

A quadrilateral is a rhombus if and only if the diagonals are perpendicular bisectors of each other.

This is an "if and only if" proof, so there are two things we have to prove:

One way to prove this is to use congruent triangles.

However, note that this essentially runs through the proof of one of the isosceles triangle theorems which we have already proved. Note the SSS reason after we established the reflexive side. We do not have to do it again. We can simply refer to it.

For the converse, 

there is the congruent tirangle proof.

But, again we are running through a proof of one of the isosceles triangle theorems. We could simply refer to it instead of proving it again.

One could also use the fact that a point is equidistant from two given points if and only if it is on the perpendicular bisector of the line segment between them.