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*}
\)
In the previous article, we proved this relationship to be exact, and gave an intuitive proof for triangles. We will now examine the full proof for triangles in Yuktibhāṣā of Jyeṣṭhadeva.
Consider triangle ABC, with BC being the base (भूमि:). The perpendicular from AP to BC (लम्बः) divides BC into two segments (आबाधौ) \(d_1\) and \(d_2\). Following modern convention, we denote the length of the side opposite A (BC) as a, AC as b and AB as c.
The Area of the triangle ABC
Ar = \frac{1}{2}a.h \\
Ar^2 = \frac{1}{4}a^2.h = (\frac{a}{2})^2.h
\)
Our aim is to show that this is equivalent to \( Ar^2 = s.(s-a).(s-b).(s-c)\)
Do do this, we first note that
\(
s.(s-a) = (\frac{b+c}{2})^2-(\frac{a}{2})^2 = (\frac{b+c}{2})^2-(\frac{d_1+d_2}{2})^2 \\
h^2 = c^2 – d_1^2 = b^2- d_2^2 = \frac{b^2+c^2}{2} – \frac{d_1^2+d_2^2}{2} \\
s.(s-a) \approx h^2 \\
(s-b)(s-c) = (\frac{a}{2})^2 – (\frac{b}{2} – \frac{c}{2})^2 \\
(s-b)(s-c) \approx (\frac{a}{2})^2 \\
s.(s-a)(s-b)(s-c) \approx (\frac{a}{2})^2.h^2 = Ar^2
\)
This proves that the two expressions are at least approximately equal. To prove exact equality, we must prove that the differences between the factors are in such proportion that their product evens out.
\(
\begin{align*}
h^2 – s.(s-a) &= (\frac{b^2+c^2}{2} – \frac{d_1^2+d_2^2}{2}) – ((\frac{b+c}{2})^2-(\frac{d_1+d_2}{2})^2) \\
&= (\frac{b-c}{2})^2-(\frac{d_2-d_1}{2})^2 \\
s.(s-a) &= h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2) \\
\end{align*}
\)
and therefore
\(s.(s-a).(s-b).(s-c) = (h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)).((\frac{a}{2})^2 – (\frac{b}{2} – \frac{c}{2})^2)\)
with one factor a bit more than \(h^2\), and another a bit less than \(a^2\).
To prove that everything evens out on multiplication, we use the relationship \( p.q = (p+r).(q – q.\frac{r}{p+r})\), which is one of the multiplication tricks used by the Kerala School. Therefore,
\(
\begin{align*}
Ar^2 &= h^2.(\frac{a}{2})^2 \\
&= (h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2))((\frac{a}{2})^2 – (\frac{a}{2})^2.\frac{((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)}{h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)}) \\
&= (h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2))((\frac{a}{2})^2 – (\frac{a}{2})^2.\frac{((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)}{((\frac{b+c}{2})^2-(\frac{d_1+d_2}{2})^2)}) \\
\end{align*}
\)
To make further progress, we use the ābādha relationship proven previously
\(
d_2^2 – d_1^2 = b^2 – c^2 \\
\frac{b-c}{d_2+d_1} = \frac{d_2-d_1}{b+c} \\
\frac{(\frac{b-c}{2})}{(\frac{d_2+d_1}{2})} = \frac{(\frac{d_2-d_1}{2})}{(\frac{b+c}{2})} \\
\frac{(\frac{b-c}{2})^2}{(\frac{d_2+d_1}{2})^2} = \frac{(\frac{d_2-d_1}{2})^2}{(\frac{b+c}{2})^2} \\
\frac{(\frac{b-c}{2})^2}{(\frac{d_2+d_1}{2})^2} = \frac{(\frac{d_2-d_1}{2})^2-(\frac{b-c}{2})^2}{(\frac{b+c}{2})^2-(\frac{d_2+d_1}{2})^2} \\
\frac{(\frac{b-c}{2})^2}{(\frac{a}{2})^2} = \frac{(\frac{d_2-d_1}{2})^2-(\frac{b-c}{2})^2}{(\frac{b+c}{2})^2-(\frac{d_2+d_1}{2})^2} \\
\)
Thus
\(
\begin{align*}
Ar^2 &= (h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2))((\frac{a}{2})^2 – (\frac{a}{2})^2.\frac{((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)}{((\frac{b+c}{2})^2-(\frac{d_1+d_2}{2})^2)}) \\
&= (h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)) ((\frac{a}{2})^2 – (\frac{a}{2})^2.\frac{(\frac{b-c}{2})^2}{(\frac{a}{2})^2}) \\
&= (h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)) ((\frac{a}{2})^2 – (\frac{b-c}{2})^2) \\
&= s.(s-a).(s-b).(s-c)
\end{align*}
\)
Which is what we set out to prove.
And so, we have yuktis to prove सर्वदोर्युतिदलं as exact for areas of cyclic quadrilaterals as well as triangles.
