Mathematics in Ancient and Medieval India c9.26

c9.26

Let PP₁ be a small increase in the arc PX (Fig. 28) and P₀ be the middle point of the small arc. The perpendicualrs PQ, P₀Q₀, P₁ Q₁ and PS are drawn. From the similar triangles P₁SP and OQ₀P₀, we have P₁S = (OQ₀ · P₁P)/OP and PS = (P₀Q₀ · P₁P)/OP₀.