Trigonometry/Area of a quadrilateral
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 = 1⁄2AC.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 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)
- a2+d2-b2-c2 = 2ad.cos(A)-2bc.cos(C) ... (i)
- S = area(BAD) + area(BCD) = 1⁄2ad.sin(A) + 1⁄2bc.sin(C)
- 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)
- cos(A+C) = cos(2α) = 2cos2(α)-1
- 16S2 = 4(ad+bc)2 - (a2+d2-b2-c2)2 - 16abcd.cos2(α).
- S2 = (σ-a)(σ-b)(σ-c)(σ-d) - abcd.cos2(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,
Thus AC.BD = ac+bd (as we already knew) and
The circumcircle of ABCD is also the circumcircle of triangle ABD, so
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. a2+d2-b2-c2 = 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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