Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Let ABCD be a parallelogram.
Let us draw perpendicular DE on extended side AB, and AF on side DC.
Applying Pythagoras theorem in ΔDEA, we obtain
DE2 + EA2 = DA2 … (i)
Applying Pythagoras theorem in ΔDEB, we obtain
DE2 + EB2 = DB2
DE2 + (EA + AB)2 = DB2
(DE2 + EA2) + AB2 + 2EA × AB = DB2
DA2 + AB2 + 2EA × AB = DB2 … (ii)
Applying Pythagoras theorem in ΔADF, we obtain
AD2 = AF2 …
To view the solution to this question please