Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squar

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.

Solution:

Let us consider, ABCD be a parallelogram. Now, draw perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.

By applying Pythagoras Theorem in ∆DEA, we get,

$DE^2 + EA^2 = DA^2 $……………….… (i)

By applying Pythagoras Theorem in ∆DEB, we get,

$DE^2 + EB^2 = DB^2$

$DE^2 + (EA + AB)^2 = DB^2$

$(DE^2 + EA^2) + AB^2 + 2EA × AB = DB^2$

$DA^2 + AB^2 + 2EA × AB = DB^2$ ……………. (ii)

By applying Pythagoras Theorem in ∆ADF, we get,

$AD^2 = AF^2 + FD^2$

Again, applying Pythagoras theorem in ∆AFC, we get,

$AC^2 = AF^2 + FC^2 = AF^2 + (DC − FD)^2$

$= AF^2 + DC^2 + FD^2 − 2DC × FD$

$= (AF^2 + FD^2) + DC^2 − 2DC × FD AC^2$

$AC^2= AD^2 + DC^2 − 2DC × FD$ ………………… (iii)

Since ABCD is a parallelogram,

AB = CD ………………….…(iv)

And BC = AD ………………. (v)

In ∆DEA and ∆ADF,

$\angle DEA = \angle AFD$ (Each 90°)

$\angle EAD = \angle ADF (EA || DF)$

AD = AD (Common Angles)

∆EAD ≅ ∆FDA (AAS congruence criterion)

EA = DF ……………… (vi)

Adding equations (i) and (iii), we get,

$DA^2 + AB^2 + 2EA × AB + AD^2 + DC^2 − 2DC × FD = DB^2 + AC^2$

$DA^2 + AB^2 + AD^2 + DC^2 + 2EA × AB − 2DC × FD = DB^2 + AC^2$

From equation (iv) and (vi),

$BC^2 + AB^2 + AD^2 + DC^2 + 2EA × AB − 2AB × EA = DB^2 + AC^2$

$AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2$

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