If it is circumscriptible then consider the contact points E,H,G,F and measure lengths ED=DH etc. to show that AB+CD=BC+DA. For the converse, assume AB + CD = BC + DA, and AB < AD. Then BC < CD, and there are points X on AD and Y on CD s.t. AX = AB and CY = BC. Then DX = DY . Let K be the circumcenter of triangle BXY . AK bisects angle A since the triangles AKX and AKB have equal sides. Similarly, CK and DK are bisectors of angles C and D respectively. It follows that K is equidistant from the sides of the quadrilateral. The quadrilateral has an incircle with center K. (Proof taken from the notes of Paul Yiu: EuclideanGeometryNotes.pdf).

Look at Circumscriptible_Construction.html for the construction of all circumscriptibles with AB+CD = BC+DA = t.

CircumscriptibleQuadrilateral.html

CircumscriptibleQuadrilateral2.html

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