As shown in the figure, the point E is on the diagonal BD of the square ABCD, and be = AB, EF ⊥ BD, EF and CD intersect at the point F to prove de = EF = FC

As shown in the figure, the point E is on the diagonal BD of the square ABCD, and be = AB, EF ⊥ BD, EF and CD intersect at the point F to prove de = EF = FC

prove
Because EF is vertical to BD, the angle def = angle DCB = 90 degrees
Because the angle BDC is the common angle, the triangle Fed is similar to the triangle BCD
Because ABCD is square, the angle DFE = angle CBD = angle BDC = 45 degrees
So EF = De
Connect CE because be = AB = BC, so angle BEC = angle BCE,
Because the angle Feb = angle FCB = 90 ° the angle FEC = angle FCE
So EF = FC
In conclusion, de = EF = FC