With the jyā recurrence without the standard radius having been demonstrated in the previous article, we return to cyclic quadrilaterals, and the proof of Bhāskarācārya's verse on the areas of cyclic quadrilaterals and triangles:

सर्वदोर्युतिदलं चतुःस्थितं बाहुभिर्विरहितं च तद्धतेः ।

मूलमस्फुटफलं चतुर्भुजे स्पष्टमेवमुदितं त्रिबाहुके ॥*Take the sum of all sides, and divide it into half. Multiply the result by the four differences of itself with each side. The square root of the result is is the approximate area of a quadrilateral, and precise for a triangle. (This is exact for a cyclic quadrilateral)*

\begin{align*}

A &= \sqrt{s.(s-a)(s-b)(s-c)(s-d)} &\textit{cyclic quadrilateral} \\

A &= \sqrt{s.(s-a)(s-b)(s-c)} &\textit{triangle} \\

\end{align*}

\)

## Yukti

Consider a cyclic quadrilateral ABCD, with sides AB=a (भूमिः – longest side), BC=b, DA=d (भुजे), and CD=c (मुखम्) (not to be confused with the cyclic diameter, which we will not need for this yukti). Here, we assume \(c > b \), and will consider the opposite case a bit later. Consider the primary diagonal BD, with midpoint at M. Draw perpendiculars (लम्बौ) AE and CF to BD. Extend AE to P and CF to Q such that AP = CQ = AE+CF.

We compute the area of ABCD (denoted as Ar) as the sum of areas of ABD and BCD, which in turn is computed as \(\frac{1}{2}.BD.(AE+CF) = \frac{1}{2}.BD.AP \) by the triangle area law (लम्बगुणं भूम्यर्धं स्पष्टं त्रिभुजे फलं भवति). Now if we can compute AP, we have what we need. Examinig the rectange APCQ, we see that

\(

\begin{align*}

AP^2 &= AC^2-PC^2 \\

&= AC^2-EF^2 \\

Ar^2 &= \frac{1}{4}.BD^2.(AC^2-EF^2) \\

\end{align*}

\)

This gets us a step further. We already know BD and AC in terms of the sides from a previous article as the square root of the ratio of the associated *कर्णाश्रितभुजाघातैक्यम्* with that of the other diagonal, multiplied by the भुजाप्रतिभुजाघातैक्यम्. Therefore

\(

BD^2 = \frac{(cd+ab)}{(ad+bc)}. (ac+bd) \\

AC^2 = \frac{(ad+bc)}{(cd+ab)}. (ac+bd)

\)

Now all we need is to find EF, which we can see from the figure as the distance between the perpendiculars AE and CF on diagonal BD, or the लम्बनिपातान्तरम् .

### Computing लम्बनिपातान्तरम्

From the figure, we can see that EF = ME + MF, and

\(

\begin{align*}

ME &= MD – DE \\

&= \frac{1}{2}(BE+ED) – ED \\

&= \frac{1}{2}(BE-ED) \\

MF &= MB – FB \\

&= \frac{1}{2}(DF+FB) – FB \\

&= \frac{1}{2}(DF – FB) \\

\end{align*}

\)

Now BE and ED are the आबाधौ of triangle ABD and DF and FB are the आबाधौ of triangle BCD. This means

\((BE-ED) = \frac{AB^2-DA^2}{BD} \\

(DF-FB) = \frac{CD^2-BC^2}{BD} \\

EF = \frac{1}{2}.\frac{(AB^2+CD^2)-(DA^2+BC^2)}{BD}

\)

AB > AD, since we've assumed AB as the longest side. We've also assumed c > b, or CD > BC. What if BC > CD? Then, the perpendiculars CF and AE will fall on the *same side* of midpoint M on BD as below.

Now:

\(EF = ME – MF \\

ME = \frac{1}{2}(BE-ED) \\

MF = \frac{1}{2}(DF – FB) \\

(BE-ED) = \frac{AB^2-DA^2}{BD} \\

(DF-FB) = \frac{BC^2-CD^2}{BD} \\

EF = \frac{1}{2}.\frac{(AB^2+CD^2)-(DA^2+BC^2)}{BD}\\

\)

Thus, we end up with the same expression for EF, irrespective of whether c > b or b > c.

\(

\begin{align*}

EF &= \frac{1}{2}.\frac{(AB^2+CD^2)-(DA^2+BC^2)}{BD} \\

&= \frac{1}{2}.\frac{(a^2+c^2)-(d^2+b^2)}{BD}

\end{align*}

\)

### Putting it all together

\(\begin {align*}

Ar^2 &= \frac{1}{4}.BD^2.(AC^2-EF^2) \\

&= \frac{1}{4}.(BD^2.AC^2 – BD^2.EF^2) \\

BD^2.AC^2 &= \frac{(cd+ab)}{(ad+bc)}. (ac+bd) . \frac{(ad+bc)}{(cd+ab)}. (ac+bd) \\

&= (ac+bd)^2 \\

BD^2.EF^2 &= BD^2 .\frac{1}{4}.\frac{((a^2+c^2)-(d^2+b^2))^2}{BD^2} \\

&= \frac{1}{4}.((a^2+c^2)-(d^2+b^2))^2 \\

Ar^2 &= \frac{1}{4}.((ac+bd)^2 – \frac{1}{4}.((a^2+c^2)-(d^2+b^2))^2) \\

&= (\frac{ac+bd}{2})^2 – (((\frac{a}{2})^2+(\frac{c}{2})^2)-((\frac{d}{2})^2+(\frac{b}{2})^2)))^2 \\

&= (\frac{ac+bd}{2} + (((\frac{a}{2})^2+(\frac{c}{2})^2)-((\frac{d}{2})^2+(\frac{b}{2})^2)))) . (\frac{ac+bd}{2} – (((\frac{a}{2})^2+(\frac{c}{2})^2)-((\frac{d}{2})^2+(\frac{b}{2})^2)))) \\

&= ((\frac{a}{2} + \frac{c}{2})^2 – (\frac{b}{2}-\frac{d}{2})^2). ((\frac{d}{2} + \frac{b}{2})^2 – (\frac{a}{2}-\frac{c}{2})^2) \\

&= \frac{a+c+b-d}{2} . \frac{a+c+d-b}{2} . \frac{d+c+b-a}{2}. \frac{d+c+a-b}{2}

\end{align*}

\)

To simplify further, we note that \(s = \frac{a+b+c+d}{2}\)

\(

\begin {align*}

Ar^2 &= \frac{a+c+b-d}{2} . \frac{a+c+d-b}{2} . \frac{d+c+b-a}{2}. \frac{d+c+a-b}{2} \\

&= (s-d) . (s-b). (s-a). (s-c) \\

Ar &= \sqrt{(s-a).(s-b).(s-c).(s-d)} \\

\end{align*}

\)

Which is सर्वदोर्युतिदलं चतुःस्थितं बाहुभिर्विरहितं च तद्धतेः मूलम्, which is where we need to be.

## Extending this yukti to a triangle

The simplest way to extend this result to a triangle, is to consider a triangle (त्र्यश्रम्) as a cyclic quadrilateral with the smallest side d = 0. This will lead directly to

\(

Ar = \sqrt{s.(s-a).(s-b).(s-c)}

\)

We also have a direct yukti for this expression, which we will see in a later article.