Brahmagupta's Formula

#### by

## Kala Fischbein and Tammy Brooks

**Prove: For a cyclic quadrilateral with sides of length a, b, c, and d,
the area is given by **

**where s is the semiperimeter. **

**Given: **

**Draw chord AC. Extend AB and CD so they meet at point P. **

**Angle ADC and Angle ABC subtend the same chord AC from the two arcs
of the circle. Therefore they are supplementary. Angle ADP is supplementary
to Angle ADC. So **

**Triangle PBC and Triangle PDA are similar. The ratio of similarity
is **

**Area of Triangle PDA = ***** Area of Triangle PBC**

Area ABCD = Area of Triangle PBC - Area of Triangle PDA

Let A = Area of quadrilateral ABCD and T = Area of Triangle PBC.

**A = T - ****T = ****T = **** T**
**Let PA = e and PD = f. Applying Heron's Formula, the area of triangle
PBC is**

**Therefore, **

**(Note: We have used s at this point for the semiperimeter of the TRIANGLE.
In what follows, we will substitute for s, e, and f in terms of a, b, c,
and d. Eventually we will return to the use of s to represent the semiperimeter
of the quadrilateral.** **)**

1. First, we want a substitution for e in terms of a, b, c, and d.

**2. Next, we want a substitution for f in terms of a, b, c, and d.**

**3. Now we will make some substitutions in the formulas for the triangle.**

**4. Now evaluate s(s - (e + a)**

**5. Likewise, we now evaluate s - (f + c) in terms of a, b, c, and d.**

**6. Now, we evaluate (s - b)**

**7. Now we are ready to evaluate the Area of the quadrilateral in terms
of a, b, c, d.**

**Therefore, **

**where s is the semiperimeter of the cyclic quadrilateral **

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Microsoft Word 6.0 document for this proof of Brahmagupta's Formula.**

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