Firstly, it is proved that the triangle ABE is congruent with the triangle ACF, so be = cf, and the perimeter EC+cf+EF = BC+EF. As shown in the title, you can calculate BC = √ 2, and find the minimum value to find the minimum value of EF, which is equal to twice the square of AE. When AE is the minimum value, AE is the minimum value on the BC side of triangle ABC, AE = √ 2/2, and EF is equal to 655.