In a right triangle ABC, angle a is a right angle, point D is the midpoint of BC, e is on AB, f is on AC, De is perpendicular to DF, and we prove that EF square = be square + CF square

In a right triangle ABC, angle a is a right angle, point D is the midpoint of BC, e is on AB, f is on AC, De is perpendicular to DF, and we prove that EF square = be square + CF square

Because angle a is a right angle and De is perpendicular to DF, the quadrilateral EDFA is a rectangle,
Because point D is the midpoint of BC, e is the midpoint of AB and F is the midpoint of AC, that is, be = De, DF = CF
And because De is perpendicular to DF, EF squared = de squared + DF squared
So EF squared = be squared + CF squared