ab = hR - can you verify that? It says that in a triangle the product of two sides equals the product of the diameter circumscribed circle and the altitude to the third side.

In a quadrilateral, you have two pairs of triangles sharing a base (the third side in 1). Bases are the diagonals of the quadrilateral. Consider one pair at a time, say, ab = ht and cd = gt, where h is the altitude to from the ab vertex to the diagonal m, I belive. g is the altitude from the cd vertex to the sam diagonal.

Each of these triangles contains a piece of the diagonal n which plays the hypotenuse, one in a triangle with side h, the other in a triangle with side g. The triangles are similar. You can write

h = cos(theta)×(one piece) and
g = cos(theta)×(the other piece).

On summing up you get

h + g = cos(theta)·n

Thus adding ab = ht and cd = gt gives you

nt·cos(theta) = ab + cd.

Sorry if I confused m and n.