Given a quadrilateral with sides a, b, c, d, diagonals e, f, and area S, we have

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

The angle V of the diagonals is given by

$\tan V={\frac{4S}{a^{2}+c^{2}-b^{2}-d^{2}}}.$

Deduce from this a solution of the preceding exercise. (Having fixed the position of one side, each of the remaining two vertices will be the intersection of two circles.)

Question

Given a quadrilateral with sides a, b, c, d, diagonals e, f, and area S, we have

[latex]4e^{2}f^{2}=(a^{2}+c^{2}-b^{2}-d^{2})^{2}+16S^{2}.[/latex]

The angle V of the diagonals is given by

[latex] an V={\frac{4S}{a^{2}+c^{2}-b^{2}-d^{2}}}.[/latex]

Deduce from this a solution of the preceding exercise. (Having fixed the position of one side, each of the remaining two vertices will be the intersection of two circles.)

Accepted Answer

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 [latex]f^{2}\,=\,B N^{2}+D N^{2}\,=\,M L^{2}+D N^{2}.[/latex] We multiply the left side by 4e² , and the right side by 4AC² , which are two equal expressions, to get:

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

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

as required.

We next compute tan [latex]\hat{{V}}[/latex], where [latex]\hat{{V}}[/latex] is the (acute) angle between the two diagonals of the quadrilateral. We have: [latex]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}.[/latex] Replacing the numerator and denominator of this fraction by the expressions derived above, we have:

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

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: [latex]B P\:=\:{\frac{a c}{d}};\:\:A C:A P\:=\:A D:A B\:=[/latex] d : a. Also, [latex]{\widehat{C A D}}={\widehat{B A P}},[/latex] which implies that [latex]\widehat{C A P}=\widehat{D A B}.[/latex] Therefore triangles ACP, ADB are also similar [latex]({\bf118},\,2^{\circ}),[/latex] and [latex]C P={\frac{e f}{d}}.[/latex]

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) [latex]\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}.[/latex]

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 = [latex]\frac{a c}{d};\;\;C P\:=\:\frac{e f}{d},[/latex] 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.

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

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