# Triangles

#### Question 6:

Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

#### Answer:

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

DE^{2}
+ EA^{2}
= DA^{2} …
(*i*)

Applying Pythagoras theorem in ΔDEB, we obtain

DE^{2}
+ EB^{2}
= DB^{2}

DE^{2}
+ (EA + AB)^{2}
= DB^{2}

(DE^{2}
+ EA^{2})
+ AB^{2}
+ 2EA × AB = DB^{2}

DA^{2}
+ AB^{2}
+ 2EA × AB = DB^{2} …
(*ii*)

Applying Pythagoras theorem in ΔADF, we obtain

AD^{2} = AF^{2}
…

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