The proof that the sum of squares of diagonals of parallelogram is equal to the sum of squares of four sides RT I hope to be more detailed. If some trigonometric functions are involved, I also hope to explain them

The proof that the sum of squares of diagonals of parallelogram is equal to the sum of squares of four sides RT I hope to be more detailed. If some trigonometric functions are involved, I also hope to explain them

Method 1: make the height of BC side through two points a and D, and the perpendicular feet are e and f respectively, then △ ade ≌ △ dcfbe = CF, AE = DF. Using Pythagorean theorem, we get BD square = BF square + DF square, BD square = (BC + CF) square + DF square = BC square + 2 * BC * CF + CF square + DF square, AC square = AE square + CE square = AE square + (bc-be) square =