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Heron's formula for the area of a triangle is the special case obtained by taking d = 0. The relationship between the general and extended form of Brahmagupta's formula is similar to how the law of cosines extends the Pythagorean theorem .
Proof of the theorem. We need to prove that AF = FD.We will prove that both AF and FD are in fact equal to FM.. To prove that AF = FM, first note that the angles FAM and CBM are equal, because they are inscribed angles that intercept the same arc of the circle (CD).
The center of area divides this segment in the ratio (when taken from the short to the long side) [21]: p. 862 a + 2 b 2 a + b . {\displaystyle {\frac {a+2b}{2a+b}}.} If the angle bisectors to angles A and B intersect at P , and the angle bisectors to angles C and D intersect at Q , then [ 19 ]
The British flag theorem can be generalized into a statement about (convex) isosceles trapezoids.More precisely for a trapezoid with parallel sides and and interior point the following equation holds:
A set of sides that can form a cyclic quadrilateral can be arranged in any of three distinct sequences each of which can form a cyclic quadrilateral of the same area in the same circumcircle (the areas being the same according to Brahmagupta's area formula). Any two of these cyclic quadrilaterals have one diagonal length in common. [17]: p. 84
The formula for the area of a trapezoid can be simplified using Pitot's theorem to get a formula for the area of a tangential trapezoid. If the bases have lengths a, b, and any one of the other two sides has length c, then the area K is given by the formula [2] (This formula can be used only in cases where the bases are parallel.)