As shown in the figure, e and F are the edges ab of square ABCD, and the point on ad ∠ ECF = 45 ° is proved to be EF = DF + be

As shown in the figure, e and F are the edges ab of square ABCD, and the point on ad ∠ ECF = 45 ° is proved to be EF = DF + be

Make an angle to the right through C, DCG = angle BCE, intersect ad extension line to g, because DC = BC, angle GDC = 90 degrees = angle B, so triangle EBC is equal to triangle GDC, so be = DG, and because angle FCE = 45 degrees, so angle FCD + angle ECB = 90 degrees - 45 degrees = 45 degrees = angle FCG, so triangle ECF and FCG are identical, so EF = FG = FD + DG = FD = be

In the square ABCD, e is a point on the edge of AB, f is a point on the edge of AD, ab = 12, EF = 10, ∠ ECF = 45 °, find the length of be?

It is easy to prove that EF = DF + be
Let be = X. (12 - (10-x)) ² + (12-x) ² = 10 & #178; [⊿ AEF Pythagorean theorem] x = 4 or 6