M is the midpoint of line AB and N is the midpoint of line AC, and T is the intersection of line BN and line CM. If line BN is perpendicular to line AC, BN=12, and AC=14, then find CT.
Since M is a median of AB and N is the median of AC, their intersection occurs (2/3) of the way from vertex B to N
So... TN is (1/3) of the distance from B to N = (1/3)(12) = 4
And TNC is a right triangle with legs NC, TN and hypotenuse CT
CT = sqrt ( NC^2 + TN^2 ) = sqrt ( 7^2 + 4^2) = sqrt ( 65)