254 dhruv raina
these demonstrations, we are reminded, is similar to Wallis’ demonstration
that appeared in the treatise on angular sections. Colebrooke sets Wallis’
and Bhaskara’s demonstrations side by side, such that Bhaskara’s method is
apprehended in Wallis’ idiom ( Figure 5.8 ). 97
Wallis Bhaskara
In a rectangular triangle, C and D
designate the sides and B the hypotenuse.
Th e segments are χ and δ.
Using the same symbols for the sides and
segments, Bhaskara’s demonstration
B : C :: C : χ B : C :: C : χ
B : D : D : δ B : D : D : δ
Th erefore Th erefore
C^2 = Bχ χ = C^2 /B
D^2 = Bδ δ = D^2 /B
Th erefore Th erefore
C^2 + D^2 =(Bχ + Bδ) = B(χ + δ) = B^2 B = χ + δ = C^2 /B + D^2 /B
B^2 = C^2 + D^2
We shall now try to illustrate Bhaskara’s procedure above as it appears
in Colebrooke’s translation, but I shall adopt a contemporary form of the
argument. Th e problem that Bhaskara poses in §146 of the Bija-Ganita is:
‘Say what is the hypotenuse in a plane fi gure, in which the side and upright
are equal to 15 and 20? And show the demonstration of the received mode
of composition’. 98 So consider a right-angled triangle ABC whose sides are
15 and 20 and rotate the fi gure as above. Drop a perpendicular to the side
AC and let AD = χ and DA = δ. Now AC is the hypotenuse of the triangle
ABC and BC and AD of triangles BCD and DBA respectively.
Figure 5.8 A right-angled triangle ABC and its height BD.
C D
B
15 20
A
97 C1817: xvi–xvii.
98 C1817: 220.