32. If a quadrilateral figure does not admit of having a circle described round it, the sum of the rectangles contained by the opposite sides is greater than the rectangle contained by the diagonals.
Let ABCD be a quadrilateral figure which does not admit of having a circle described round it; then the rectangle AB, DC, together with the rectangle BC, AD, shall be greater than the rectangle AC, BD.
For, make the angle ABE equal to the angle DBC, and the angle BAE equal to the angle BDC; then the triangle ABE is similar to the triangle BDC (VI. 4); therefore AB is to AE as DB is to DC; and therefore the rectangle AB, DC is equal to the rectangle AE, DB.
Join EC. Then, since the angle ABE is equal to the angle DBC, the angle CBE is equal to the angle DBA. And because the triangles ABE and DBC are similar, AB is to DB as BE is to BC; therefore the triangles ABD and EBC are similar (VI. 6); therefore CB is to CE as DB is to DA; and therefore the rectangle CB, DA is equal to the rectangle CE, DB.
Therefore the rectangle AB, DC, together with the rectangle BC, AD is equal to the rectangle AE, BD together with the rectangle CE, BD; that is, equal to the rectangle contained by BD and the sum of AE and EC. But the sum of AE and EC is greater than AC (I. 20); therefore the rectangle AB, DC, together with the rectangle BC, AD is greater than the rectangle AC, BD.
