As shown in the figure, take a point E on the diagonal BD of square ABCD, make be = BC, make ef vertical BD, intersect CD with F. the line segment De is equal to EF and FC? Why?

In square ABCD, BD bisects ADC,
So, EDC = 45,
And EF vertical BD,
So delta DEF is an isosceles right triangle,
So de = EF,
Even BF,
Because be = BC
BF common edge
So △ bef ≌ △ BCF
So EF = FC
So de = EF = FC

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