## viernes, 7 de mayo de 2021

### Using the half-angle formulas to derive Mahavira's identities

In a cyclic quadrilateral $ABCD$, let $a$, $b$, $c$, $d$ denote the lengths of sides $AB$, $BC$, $CD$, $DA$, and $m$, $n$ the lengths of the diagonals $BD$ and $BC$. Then Mahavira's result is expressed as
$$m^2=\frac{(ab+cd)(ac+bd)}{ad+bc}\tag{1}$$
$$n^2=\frac{(ac+bd)(ad+bc)}{ab+cd}\tag{2}.$$

Proof. By the Law of Cosines,

\begin{align*}m^2&=a^2+d^2-2ad\cos{A}\\&=a^2+d^2-2ad(2\cos^2{\frac{A}{2}}-1)\\&=(a+d)^2-4ad\cos^2{\frac{A}{2}}\end{align*}

Substituting from the half-angle formula (see formula $(5)$ in this page) we get

\begin{align*}m^2&=(a+d)^2-\frac{ad[(a+d)^2-(b-c)^2]}{ad+bc}\\&=\frac{bc(a+d)^2+ad(b-c)^2}{ad+bc}\\&=\frac{a^2bc+bcd^2+ab^2d+ac^2d}{ad+bc}\\&=\frac{(ab+cd)(ac+bd)}{ad+bc}.\end{align*}
$\square$

Similarly we can get $(2)$.

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