In the previous article in this series, we began a search for a jyā recurrence without using the standard radius R, and went off into triangles and cyclic quadrilaterals. In this article, we will dive deeper into the latter.

## The sampurṇajyā relationship

First, we consider an relationship about sampurṇajyā and associated arcs (चाप) in a circle. This will be very useful for us in the following discussion.

Consider a वृत्तान्तर्गतचतुरश्रम् (cyclic quadrilateral). By convention, we consider AB, the longest side on the west (ie, the bottom of the picture). This is called the भूमिः or base. The sides AD and BC are called भुजे (singular – भुजा), and the side DC is the मुखम् . We draw DM as the लम्बः (perpendicular) to AC, a कर्णः (diagonal). EP is parallel to DM and of the same length, so AE = DC. Therefore, arc ED = Arc AD – Arc AE = Arc AD – Arc DC. Also, ED = PM = AM-AP = AM-MC (since arcs AE and DC are the same)

In the previous article, we had noted the relationship between the आबाधौ (basal segments) of a triangle, which in the case of triangle ADC leads us to \(AD^2-DC^2 = AM^2 – MC^2 = AC.(AM-MC)\). Segments AD and DC are the full-chords (सम्पूर्णज्या) of arcs AD and DC. Similarly, AC is the full-chord of arc AC the sum of arcs AM and MC, and AM-MC = ED = PM is the full-chord corresponding to the difference of arcs AM and MC.

Thus, setting

s_1 = arc(AD) \\

s_2 = arc(DC)

\)

leads to the first sampurṇajyā relationship

\([sampurṇajyā(s1)]^2 – [sampurṇajyā(s2)]^2 = sampurṇajyā(s1).sampurṇajyā(s2)

\)

Since the ADC can be arbitrary, we can also set

s_1 = arc(AC) \\

s_2 = arc(AM-MC) = arc(ED)

\)

which leads to the second sampurṇajyā relationship

\(sampurṇajyā(s_1).sampurṇajyā(s_2) = [sampurṇajyā(\frac{s_1+s_2}{2})]^2 – [sampurṇajyā(\frac{s_1-s_2}{2})]^2

\)

## Diagonals of a वृत्तान्तर्गतचतुरश्रम् (cyclic quadrilateral)

Consider the rather imposing figure of a cyclic quadrilateral ABCD, as above. AB is the भूमिः (longest side) and sides AD, BC are the भुजे.

What about all the other stuff then? Points E and F on the circle are marked such that AE = DC and AF = BC (and so are the corresponding arcs). We connect the midpoints of Arcs ED and FB, called H and G respectively, to create a diameter HG. (*Why is HG a diameter? Ponder the symmetry of AEDCBF*).

### त्रयः कर्णाः – Three diagonals

BD is the first diagonal, corresponding to the sampurṇajyā of arc(BD) (which is equal to arc(BC) + arc(CD)). Similarly AB, the second diagonal is the sampurṇajyā of arc(AB), which is the sum of arc(AC) + arc(CB).

Let us now consider DF. DF is sampurṇajyā of \(arc(DF) = arc(AD) + arc(AF) = arc(AD) + arc(AC)\), ie – the sampurṇajyā of the sum of the arcs of opposite sides! *(what about EB then? By symmetry, EB=DF)*

**Thus, we conceptualize DF (=EB) as the third diagonal of the वृत्तान्तर्गतचतुरश्रम्, corresponding to the sampurṇajyā of the sum of arcs AD and BC (or arcs AB and CD). **We think of sides, as well as diagonals (all three!) as sampurṇajyā of arcs corresponding to sides or sum of adjacent or opposite sides. This is an interesting way of thinking about cyclic quadrilaterals.

### कर्णाश्रितभुजाघातैक्यम्

That imposing, yet compact phrase means sum of products of corresponding sides associated with a diagonal. For the first diagonal DB, the pairs of corresponding sides are (AD, DC) and (AB, BC). Thus the कर्णाश्रितभुजाघातैक्यम् is \(AD.DC + AB.BC\). Using the sampurṇajyā relations, we can say

\(

AD.DC = [sampurṇajyā(\frac{arc(AD)+arc(DC)}{2})]^2 – [sampurṇajyā(\frac{arc(AD)-arc(DC)}{2})]^2 \\

AB.BC = [sampurṇajyā(\frac{arc(AB)+arc(BC)}{2})]^2 – [sampurṇajyā(\frac{arc(AB)-arc(BC)}{2})]^2

\)

Consider the sum of first terms of the two equations. We know that

\(

arc(AD)+arc(DC) = 360 – (arc(AB)+arc(BC))\\

\frac{arc(AD)+arc(DC)}{2} = 180 – \frac{arc(AB)+arc(BC)}{2}

\)

This makes sampurṇajyā of \(\frac{arc(AD)+arc(DC)}{2}\) and \(\frac{arc(AB)+arc(BC)}{2}\) jyā and koṭi of a right triangle, which means the square of their sum will be equal to the square of the diameter.

\(

[sampurṇajyā(\frac{arc(AD)+arc(DC)}{2})]^2 + [sampurṇajyā(\frac{arc(AB)+arc(BC)}{2})]^2 = d^2 \\

AD.DC + AB.BC = d^2 – [sampurṇajyā(\frac{arc(AD)-arc(DC)}{2})]^2 – [sampurṇajyā(\frac{arc(AB)-arc(BC)}{2})]^2

\)

We have simplified a bit, but have some more way to go. To simplify further, we note that

\(

\frac{arc(AD)-arc(DC)}{2} = \frac{arc(DE)}{2} = arc(DH)\\

\frac{arc(AB)-arc(BC)}{2} = \frac{arc(BF)}{2} = arc(GB) \\

\)

\(

\begin{align*}

AD.DC + AB.BC &= d^2 – [sampurṇajyā(\frac{arc(AD)-arc(DC)}{2})]^2 – [sampurṇajyā(\frac{arc(AB)-arc(BC)}{2})]^2 \\

&= d^2 – [sampurṇajyā(arc(DH)]^2 – [sampurṇajyā(arc(GB))]^2 \\

&= d^2 – DH^2 – GB^2 \\

&= GD^2 – GB^2

\end{align*}

\)

since GDH is inscribed in a semicircle, and hence a right triangle.

Again, by the sampurṇajyā relation

\(

\begin{align*}

AD.DC + AB.BC &= GD^2 – GB^2 \\

&= sampurṇajyā(arc(GD)+arc(GB)).sampurṇajyā(arc(GD)-arc(GB)) \\

&= sampurṇajyā(arc(GD)+arc(GF)).sampurṇajyā(arc(GD)-arc(GB)) \\

&= sampurṇajyā(arc(GF)).sampurṇajyā(arc(DB)) \\

&= DF . DB

\end{align*}

\)

Thus, the कर्णाश्रितभुजाघातैक्यम् – the sum of products of sides associated with a diagonal DB is the product of the diagonal (DB) with the third diagonal (DF).

A similar argument for diagonal AC will produce a corresponding result

\( AB.AD + BC.DC = AC. DF \)

Also, by a similar argument, we can prove that the भुजाप्रतिभुजाघातैक्यम् (sums of products of opposite sides) is the product of the first two diagonals

\( AB.DC + BC.AD = AC. BD \)

(*This can also be seen by considering the quadrilateral DAFC and noting that AF = BC and FC = AB. We can see that BD is the third diagonal of this quadrilateral DAFC. Considering the कर्णाश्रितभुजाघातैक्यम् of its second diagonal AC as above will quickly lead to the above result.* )

### कर्णाः

This lets us write down the diagonals in terms of the sides thus:

\(

DB = \sqrt(\frac{(AD.DC + AB.BC).(AB.DC + BC.AD)}{AB.AD + BC.DC})) \\

AC = \sqrt(\frac{(AB.AD + BC.DC).(AB.DC + BC.AD)}{AD.DC + AB.BC})) \\

DF = \sqrt(\frac{(AB.AD + BC.DC).(AD.DC + AB.BC)}{AB.DC + BC.AD})) \\

\)

The first and second diagonals are the square roots product of their कर्णाश्रितभुजाघातैक्यम् divided by that of the other and multiplied by the भुजाप्रतिभुजाघातैक्यम्. The third diagonal is the product of the two कर्णाश्रितभुजाघातैक्ये divided by the भुजाप्रतिभुजाघातैक्यम्.

### Area of the वृत्तान्तर्गतचतुरश्रम्

We can now come up with an expression for the area of a cyclic quadrilateral. First we note that adding the areas of ABC and ADC will get us what we want. That gives us the expression

\(

\begin{align*}

A &= \frac{1}{2}(AC.DM + AC.MN) \\

&= \frac{1}{2}(AC.DO)

\end{align*}

\)*(O is obtained by extending DM and BF till they intersect. Since DM is perpendicular to AC which is parallel to BF, DO Is perpendicular to BF)*

Using the relationship that the product of two sides of a triangle is equivalent to the third multiplied by its circumdiameter (which we will prove later), we see that:

\begin{align*}

DO &= DB.DF/d \\

A &= \frac{1}{2}(AC.DO) \\

&= \frac{AC.DB.DF}{2d}

\end{align*}

\)

The area of a cyclic quadrilateral is the product of its three diagonals divided by twice the circumdiameter. This is an interesting result, and a good place to stop for now, but is not totally satisfactory, since we are yet to determine the circumdiameter as a function of the sides. We have some more work to do till we prove the सर्वदोर्युतिदलं formula, which is our goal.

## Summary

### सम्पूर्णज्या relationships

\([sampurṇajyā(s1)]^2 – [sampurṇajyā(s2)]^2 = sampurṇajyā(s1).sampurṇajyā(s2) \\

sampurṇajyā(s_1).sampurṇajyā(s_2) = [sampurṇajyā(\frac{s_1+s_2}{2})]^2 – [sampurṇajyā(\frac{s_1-s_2}{2})]^2

\)

### Third diagonal of a cyclic quadrilateral

In a cyclic quadrilateral ABCD, the कर्णाश्रितभुजाघातैक्यम् (sum of product of corresponding sides) of a diagonal is the product of the diagonal and the third diagonal DF.

\( AB.AD + BC.DC = AC. DF \)

### Diagonals of a cyclic quadrilateral

\(DB = \sqrt(\frac{(AD.DC + AB.BC).(AB.DC + BC.AD)}{AB.AD + BC.DC})) \\

AC = \sqrt(\frac{(AB.AD + BC.DC).(AB.DC + BC.AD)}{AD.DC + AB.BC})) \\

DF = \sqrt(\frac{(AB.AD + BC.DC).(AD.DC + AB.BC)}{AB.DC + BC.AD})) \\

\)

### Area of a cyclic quadrilateral

\(A = \frac{AC.DB.DF}{2d}

\)

Where d is the circumdiameter of the cyclic quadrilateral.