Can anyone please proof the Brahmagupta formula for finding the area of a cyclic quadrilateral : √(s-a)(s-b)(s-c)(s-d)
Answers
Brahmagupta's Formula
Brahmagupta's formula, a special formula for finding area, is one of many marvelous intellectual accomplishments of a 7th century astronomer and mathematician from India. His name was Brahmagupta and this is his formula:

where

Let's prove Brahmagupta's formula and put it to use.
Getting Ready
Any figure with four sides is called a quadrilateral. 'Quad' means four, and 'lateral' means side. If you fit a quadrilateral inside a circle, and all four corners touch the circumference, you have a quadrilateral inscribed in a circle, also known as a cyclic quadrilateral.
See how this quadrilateral is inscribed in a circle?
a, b, c and d are the sides. These sides are the only information needed for calculating the area of an inscribed quadrilateral. S, is obtained from a, b, c and d.
Remember, the additional detail and the math that follows are only there for the derivation. The formula itself is simple: knowing the lengths of the four sides gives us the area of an inscribed quadrilateral.
A dashed line connecting opposite corners and some angles
See how x splits the quadrilateral into two triangles? See how the angles α and β are opposite angles?
The law of cosines relates the three sides of any triangle. If the angle, α is 90o, then x2= a2 + b2 (Pythagoras' theorem). If the angle is not 90o, then

What about the other triangle?

Both equations are equal to x2. Thus,

Opposite angles are supplementary in a cyclic quadrilateral. Meaning,

Using the Fundamental Facts
Since α + β = 180o, β = 180o - α.
Which says,

Expand using the cosine of the difference of two angles:

But cos 180o = -1 and sin 180o = 0:

Simplifying

Remember how

replace cos β with - cos α

Group cos α terms and factor:

A triangle with sides, a and b, subtending an angle α has an area of (1/2) ab sin α. With two triangles, the total area is

Set sin β = sin (180o - α):

Expand the sine of the difference of two angles:

Sin 180o = 0 and cos 180o = -1:

Simplifying

Thus,

replacing sin β with sin α and combining terms

Making sin α the subject of the equation:

This is an expression with sin α. Earlier,

These two expressions can be related by using a ''trick'' involving a right triangle. The side opposite the angle α is 2A while the hypotenuse is ab + cd:
The sine is the opposite, 2A, over the hypotenuse, ab+cd
The square root term comes from Pythagoras theorem.
We now have a way to define cos α. Cosine is the adjacent side divided by the hypotenuse. Referring to our right triangle:

Returning to the expression for cos α:

Substitute for cos α:

Cancelling the ab + cd terms:

Squaring both sides:

Expanding, gives -16A2 on the right-hand side. Isolating the 16A2 on the left-hand side:

Remember the formula for the difference of squares? In general, for u2 - v2:

Use the difference of squares formula to get

Interestingly:

which can be re-arranged as

And,

re-arranges to

Thus,

Once again, a difference of squares:

We saw S earlier:

Multiplying both sides by 2:

and

and

and

Substitute into our 16A2 equation:

Simplifying:

Factoring out the 2's (there are four of them),
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