In your figure AB must equal AD in length, and DC must equal CB in length so that you get a kite. If you join D to B you have two isosceles triangles with common side DB, and the triangles are DAB and DCB.
If you drop a perpendicular from vertex A to DB, DB will be bisected, and the angle at vertex A will also be bisected. Similarly DB will be bisected, and the angle at vertex C will be bisected, if you drop a perpendicular from C to DB. If X is the midpoint of DB, then AXC is a straight line perpendicular to DB, but AC and DB are the diagonals of the kite so the diagonals of a kite intersect at right angles.
The coords of A must be (B/2,A) and C must be (B/2,0). This means you can’t use C(C,0) as an arbitrary point, or you will not get a kite.
The slope of a vertical line is infinite (AC is a vertical line). The slope of a horizontal line is zero. You cannot use mm'=-1 to find the slope m' of a perpendicular to a line with slope m, if m is infinite or zero.
If you regraph the kite so that one of its sides is horizontal then you will find you can use mm'=-1 when you apply it to the slopes of the diagonals.
In the picture, the black quadrilateral ABCD is a kite, but the red quadrilateral ABCD is not a kite, and its perpendicular diagonals are not bisectors of any sort. The red vertex C is arbitrarily placed, whereas the black vertex C is placed so that black ABCD is a kite.
