Trigonometry/Area of a quadrilateral

Take any quadrilateral ABCD. Write AB=a, BC=b, CD=c, DA=a; &sigma; = $1/2$(a+b+c+d); area of ABCD = S.

Note that a+b+c-d = 2(&sigma;-d) and similarly for the other sides.

The diagonals of ABCD are AC and BD. Let the angle between them be &theta;. Then S = $1/2$AC.BD.sin(&theta;).

If ABCD is convex and the diagonals intersect at P, this is easily proved by considering the four triangles ABP, BCP, CDP, DAP since S is the sum of the areas of these four triangles. IF ABCD is not convex, then one of the vertices, say C, must lie inside the triangle ABD. We then find S as the area of ABD less the area of BCD.

Let angle A+C = 2&alpha;. To find S in terms of the sides and alpha;.

We can find BD2 by applying the cosine theorem to either of the triangles BAD, BCD. This means that


 * a2+d2-2ad.cos(A) = b2+c2-2bc.cos(C)

so


 * a2+d2-b2-c2 = 2ad.cos(A)-2bc.cos(C) ... (i)

Also


 * S = area(BAD) + area(BCD) = $1/2$ad.sin(A) + $1/2$bc.sin(C)

so


 * 4S = 2ad.sin(A) + 2bc.sin(C) ... (ii)

Taking (ii)2 + (i)2,


 * 16S2 + (a2+d2-b2-c2)2 = 4a2d2 + 4b2c2 - 8abcd.cos(A+C)

But


 * cos(A+C) = cos(2&alpha;) = 2cos2(&alpha;)-1

so


 * 16S2 = 4(ad+bc)2 - (a2+d2-b2-c2)2 - 16abcd.cos2(&alpha;).

Simplifying,


 * S2 = (&sigma;-a)(&sigma;-b)(&sigma;-c)(&sigma;-d) - abcd.cos2(&alpha;).

This expression becomes even simpler for a cyclic quadrilateral, because then cos(&alpha;) = 0 so the last term disappears.

The diagonals and circumradius of a cyclic quadrilateral
In the expression (i) above, for a cyclic quadrilateral cos(C) = -cos(A), so


 * $$2(ad+bc)\cos(A) = a^2+d^2-b^2-c^2$$.

From the cosine theorem,


 * $$BD^2 = a^2+d^2-2ad.\cos(A) = a^2+d^2 - \frac{ad(a^2+d^2-b^2-c^2)}{ad+bc} = \frac{(ab+cd)(ac+bd)}{ad+bc}$$.

Similarly,


 * $$AC^2 = \frac{(ab+cd)(ad+bc)}{ab+cd}$$.

Thus AC.BD = ac+bd (as we already knew) and


 * $$\frac{AC}{BD} = \frac{ad+bc}{ab+cd)}$$.

The circumcircle of ABCD is also the circumcircle of triangle ABD, so


 * $$R = \frac{BD}{2\sin(A)} = \frac{(ad+bc)BD}{2(ad+bc)\sin(A)} = \frac{(ad+bc)BD}{4S} = \frac{1}{4S}\sqrt{(ab+cd)(ac+bd)(ad+bc)}$$.


 * $$\tan^2 \left( \frac{B}{2} \right) = \frac{(\sigma-a)(\sigma-b)}{(\sigma-c)(\sigma-d)}$$.

Further results
If a quadrilateral is circumcyclic, i.e. such that a circle can be inscribed in it touching all four sides, then a+c=b+d. This is easily proved by noting that the lengths of the two tangents from a point to a circle are equal in length.

The radius of the inscribed circle is called the inradius and equals S/&sigma;.

Theorem: If a quadrilateral ABCD is both cyclic and circumcyclic, then


 * $$\tan^2 \left(\frac{A}{2} \right) = \frac{bc}{ad}$$.

Proof: Since ABCD is cyclic,


 * $$\cos(A) = \frac{a^2+d^2-b^2-c^2}{2(ad+bc)}$$.

Since a+c=b+d, a-d=b-c, i.e. a2+d2-b2-c2 = 2(ad-bc). Thus


 * $$\cos(A) = \frac{ad-bc}{ad+bc}$$


 * $$\tan^2 \left(\frac{A}{2} \right) = \frac{1-\cos(A)}{1+\cos(A)} = \frac{bc}{ad}$$.

If a quadrilateral ABCD is both cyclic and circumcyclic, then its area is √(abcd) and inradius is 2√(abcd)/(a+b+c+d).