Take any quadrilateral ABCD. Write AB=a, BC=b, CD=c, DA=a; σ = ^{1}⁄_{2}(a+b+c+d); area of ABCD = S.

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

The diagonals of ABCD are AC and BD. Let the angle between them be θ. Then S = ^{1}⁄_{2}AC.BD.sin(θ).

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 = 2α. To find S in terms of the sides and alpha;.

We can find BD^{2} by applying the cosine theorem to either of the triangles BAD, BCD. This means that

- a
^{2}+d^{2}-2ad.cos(A) = b^{2}+c^{2}-2bc.cos(C)

so

- a
^{2}+d^{2}-b^{2}-c^{2}= 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},

- 16S
^{2}+ (a^{2}+d^{2}-b^{2}-c^{2})^{2}= 4a^{2}d^{2}+ 4b^{2}c^{2}- 8abcd.cos(A+C)

But

- cos(A+C) = cos(2α) = 2cos
^{2}(α)-1

so

- 16S
^{2}= 4(ad+bc)^{2}- (a^{2}+d^{2}-b^{2}-c^{2})^{2}- 16abcd.cos^{2}(α).

Simplifying,

- S
^{2}= (σ-a)(σ-b)(σ-c)(σ-d) - abcd.cos^{2}(α).

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

## The diagonals and circumradius of a cyclic quadrilateralEdit

In the expression (i) above, for a cyclic quadrilateral cos(C) = -cos(A), so

- .

From the cosine theorem,

- .

Similarly,

- .

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

- .

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

- .

- .

## Further resultsEdit

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/σ.

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

- .

**Proof:** Since ABCD is cyclic,

- .

Since a+c=b+d, a-d=b-c, i.e. a^{2}+d^{2}-b^{2}-c^{2} = 2(ad-bc). Thus

- .

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

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