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Homeproof of Brahmagupta's formula

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# proof of Brahmagupta’s formula

We shall prove that the area of a cyclic quadrilateral with sides $p,q,r,s$ is given by

$\sqrt{(T-p)(T-q)(T-r)(T-s)}$ |

where $T=\frac{p+q+r+s}{2}.$

Area of the cyclic quadrilateral = Area of $\triangle ADB+$ Area of $\triangle BDC.$

$=\frac{1}{2}pq\sin A+\frac{1}{2}rs\sin C$ |

But since $ABCD$ is a cyclic quadrilateral, $\angle DAB=180^{\circ}-\angle DCB.$ Hence $\sin A=\sin C.$ Therefore area now is

$Area=\frac{1}{2}pq\sin A+\frac{1}{2}rs\sin A$ |

$(Area)^{2}=\frac{1}{4}\sin^{2}A(pq+rs)^{2}$ |

$4(Area)^{2}=(1-\cos^{2}A)(pq+rs)^{2}$ |

$4(Area)^{2}=(pq+rs)^{2}-cos^{2}A(pq+rs)^{2}$ |

Applying cosines law for $\triangle ADB$ and $\triangle BDC$ and equating the expressions for side $DB,$ we have

$p^{2}+q^{2}-2pq\cos A=r^{2}+s^{2}-2rs\cos C$ |

Substituting $\cos C=-\cos A$ (since angles $A$ and $C$ are suppplementary) and rearranging, we have

$2\cos A(pq+rs)=p^{2}+q^{2}-r^{2}-s^{2}$ |

substituting this in the equation for area,

$4(Area)^{2}=(pq+rs)^{2}-\frac{1}{4}(p^{2}+q^{2}-r^{2}-s^{2})^{2}$ |

$16(Area)^{2}=4(pq+rs)^{2}-(p^{2}+q^{2}-r^{2}-s^{2})^{2}$ |

which is of the form $a^{2}-b^{2}$ and hence can be written in the form $(a+b)(a-b)$ as

$(2(pq+rs)+p^{2}+q^{2}-r^{2}-s^{2})(2(pq+rs)-p^{2}-q^{2}+r^{2}+s^{2})$ |

$=((p+q)^{2}-(r-s)^{2})((r+s)^{2}-(p-q)^{2})$ |

$=(p+q+r-s)(p+q+s-r)(p+r+s-q)(q+r+s-p)$ |

Introducing $T=\frac{p+q+r+s}{2},$

$16(Area)^{2}=16(T-p)(T-q)(T-r)(T-s)$ |

Taking square root, we get

$Area=\sqrt{(T-p)(T-q)(T-r)(T-s)}$ |

## Mathematics Subject Classification

51-00*no label found*

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