Question P.417b: Given a quadrilateral with sides a, b, c, d, diagonals e, f,......

We use figure t417a, letting AB = a, BC = b, CD = c, DA = d, AC = e, BD = f. Let E be the intersection of the diagonals, and let L, M be the projections of vertices B, D on line AC. Finally, let N be the projection of vertex B on line DM.

From right triangle BDN we have f^{2}\,=\,B N^{2}+D N^{2}\,=\,M L^{2}+D N^{2}. We multiply the left side by 4e² , and the right side by 4AC² , which are two equal expressions, to get:

(1a)          4e^{2}f^{2}=(2A C\cdot M L)^{2}+(2A C\cdot D N)^{2}.

We work on the right hand side of this equation term by term. From equation (1) in the solution to exercise 417, we have 2AC ·ML = AB²+CD²−BC²−AD² = a²+c²−b²−d² . Also, 2AC·DN = 2AC·DM +2AC·BL = 4|ACD|+4|ABC| = 4S (using absolute value to denote areas). Substituting these expressions into (1a), we get:

(2a)           4e^{2}f^{2}=(a^{2}+c^{2}-b^{2}-d^{2})^{2}+16S^{2},

as required.

We next compute tan \hat{{V}}, where \hat{{V}} is the (acute) angle between the two diagonals of the quadrilateral. We have: t a n\ \hat{V}=t a n\ \widehat{{D}{B}N}=\frac{D N}{N B}=\frac{D N}{M L}=\frac{2A C.D N}{2A C.M L}.  Replacing the numerator and denominator of this fraction by the expressions derived above, we have:

t a n\:\hat{V}=\frac{4S}{a^{2}+c^{2}-b^{2}-d^{2}},

as required.
We now address the task of exercise 417: to construct a quadrilateral, knowing its four sides and its area. In figure t417bi, point P is chosen so that triangle ABP is similar to triangle ADC. Then we have: B P\:=\:{\frac{a c}{d}};\:\:A C:A P\:=\:A D:A B\:= d : a. Also, {\widehat{C A D}}={\widehat{B A P}}, which implies that \widehat{C A P}=\widehat{D A B}. Therefore triangles ACP, ADB are also similar ({\bf118},\,2^{\circ}), and C P={\frac{e f}{d}}.

We now can compute CP from the given sides and area of the quadrilateral. Indeed, if we divide both members of (2a) by 4d², we find:

(3a)         \left(\frac{e f}{d}\right)^{2}=\frac{1}{4}\left(\frac{a^{2}}{d}+\frac{c^{2}}{d}-\frac{b^{2}}{d}-d\right)^{2}+\left(\frac{2S}{d}\right)^{2}.

Let us fix the position of side BC (whose length is given) on the plane. Then from our discussion, we know the distances from point P to B and to C: BP = \frac{a c}{d};\;\;C P\:=\:\frac{e f}{d}, and the last is given by (3a). These two pieces of information determine point P. We can determine point A by finding the intersection of two circles: (i) the circle which is the locus of points A such that AC : AP = d : a (116), and (ii) the circle with center B and radius a. Finally, we can locate vertex D because we know its distance c from point C and its distance d from point A. This allows us to construct the required quadrilateral.