Question P.418: Construct a cyclic quadrilateral knowing its sides....
Construct a cyclic quadrilateral knowing its sides.
Suppose (fig. t418) ABCD is the required quadrilateral, and let P be a point on the plane such that triangle ABP is similar to triangle ADC.
First note that points P, B, C are collinear, because {\widehat{{A B P}}}+{\widehat{{A B C}}}={\widehat{{A D C}}}+ {\widehat{{A B C}}}=180^{\circ}. Also, from similar triangles ABP, ADC, we have B P={\frac{A B\cdot C D}{A D}}. If we fix side BC, this gives us the length of BP in terms of the given sides, and thus allows us to locate point P.
From the same similar triangles, we have AC : AP = AD : AB. Thus point A lies on the intersection of two circles: (i) the locus of points whose distances to C and P are in the ratio AD : AB (116), and (ii) a circle centered at B with radius AB (which is given). Finally, we can locate vertex D from its given distances CD, AD from vertices C and A.
