Square ABCD has a side length of 1. Point E lies on the interior of ABCD, and is on the line AC such that the length of AE is 1. Find the shortest distance from point E to a side of square ABCD
it is very confusing
On the diagonal AC, find point E such that AE = 1.
Find point X on BC such that EX is perpendicular to CB.
Since AC is a diagonal of a square whose sides are each 1, the length of AC = sqrt(2).
Since AE = 1, EC = sqrt(2) - 1..
Triangle(ACB) is similar to triangle(ECX) [by AA].
This makes CE/CA = EX/AB
---> (sqrt(2) - 1) / sqrt(2) = EX / 1
---> (sqrt(2) - 1) / sqrt(2) = EX
---> EX = 0.293 (approximately)
[It would be the same istance to side CD.]