Page 3 - LN
P. 3
Step 2: With B as centre, draw (in the interior of <A ) an arc whose radius is more than half the
length BC.
Step 3: With the same radius and with C as centre, draw another arc in the interior of <A . Let
the two arcs intersect at D. Join AD.
Conclusion: Therefore, AD is the required bisector of <A .
