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i think its bisect.
No actually the question is - in an equilateral ∆ABC, the side BC is trisected at D. Prove that 9AD^2= 7AB^2


Trisect an angle is to divide the angle into three equal halves
for e.g. if AD trisects <A and <A = 120° then it means that each of the small angle formed is 40° each that is 120°/3
1 5 1
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