Segment AN is an altitude of right ΔABC with a right angle at A. If AB = 2 sqrt5 in and NC = 1 in, find BN, AN, AC.

According to the right triangle altitude theorem, AN²=CN*NB=1*NB=NB

ΔANB is a right triangle, so AN²=AB²-NB² = (2√5) ²-NB²=20-NB²

NB=20-NB²

solve for NB: NB²+NB-20=0, (NB+5) (NB-4) = 0, NB=4 or NB=-5 (ignore this negative root)

AN²=NB=4, so AN=2

AC²=AN²+CN², AC²=2²+1², AC=√5