By Bibhutibhushan Datta
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Additional resources for Ancient Hindu Geometry: The Science Of The Sulba
This exercise can be taken as the basis for a nonstandard construction which trisects an angle: simply reconstruct Figure t49 with the given angle in place of AOE. It is known that the trisection of an arbitrary angle is not possible using classical Euclidean methods. The construction suggested here is also not possible. If we take the vertex of the given angle as O and draw any circle centered at O, we will determine point E and diameter AB. We must now ﬁnd point C along line AB so that CD = OA.
SOLUTIONS FOR BOOK I to AA and BB , respectively (as shown in the diagram). Line OF , perpendicular to DE, will pass through the intersection point of AA and BB since OF is the third altitude of the triangle formed by EA , DB , and DE. This construction works only if we assume that points A and B, as well as segment DE, lie within the limits of the diagram. Students can draw for themselves the situation in which O lies outside the angle formed by lines AA , BB . The construction remains valid in this case.
The length 1. SOLUTIONS FOR BOOK I 23 B B' X M' M p Y N' N q A Figure t32 of the path AN M B is equal to AN + N B plus the length of XY , while the length of path AN M B is equal to AN + N B = AB plus the length of XY . But the triangle inequality tells us that AN + N B > AB , so the path AM N B is shorter. Notes. As further exercise, students may be asked to show that one obtains just the same solution if we translate A (by Y X) rather than B. Students may also be asked to prove the analogous statement for two pairs of parallel lines and two given directions.