Prove that the sum of squares of four sides of a parallelogram is equal to the sum of squares of diagonals

Known;: parallelogram ABCD
Verification: AC & # 178; + BD & # 178; = AB & # 178; + BC & # 178; + CD & # 178; + Da & # 178;
It is proved that the high AE, DF
In parallelogram, ab = DC, ad = BC, ab ‖ DC
∴∠ABE=∠DCF
∵∠AEB=∠DFC=90°
∴△ABE≌△DCF
∴AE=DF BE=CF
∵AC²=AE²+EC²=AE²+(BC-BE)²=AE²+BC²-2BC×BE+BE²
BD²=DF²+BF²=AE²+(BC+CF)²=AE²+BC²+2BC×BE+BE²
∴AC²+BD²=2(AE²+BE²)+2BC²=2AB²+2BC²=AB²+BC²+CD²+DA²

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