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

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 $$\triangle$$ DEA, we get,

$$DE^2 + EA^2 = DA^2$$ ……………….… (i)

By applying Pythagoras Theorem in $$\triangle$$ DEB, we get,
$$\Rightarrow DE^2 + EB^2 = DB^2$$
$$\Rightarrow DE^2 + (EA + AB)^2 = DB^2$$
$$\Rightarrow(DE^2 + EA^2 ) + AB^2 + 2EA × AB = DB^2$$
$$\Rightarrow DA^2 + AB^2 + 2EA × AB = DB^2$$ ………. (ii)

By applying Pythagoras Theorem in $$\triangle$$ ADF, we get,
$$AD^2 = AF^2 + FD^2$$

Again, applying Pythagoras theorem in $$\triangle$$ AFC, we get,
$$AC^2 = AF^2 + FC^2$$
$$= AF^2 + (DC - FD)^2$$
$$= AF^2 + DC^2 + FD^2 - 2DC × FD$$
$$= (AF^2 + FD^2 ) + DC^2 - 2DC × FD (AC^2)$$
$$\Rightarrow AC^2= AD^2 + DC^2 - 2DC × FD$$ ………………… (iii)

Since ABCD is a parallelogram,
AB = CD ………………….…(iv)
And BC = AD ………………. (v)

In $$\triangle$$ DEA and $$\triangle$$ ADF,
$$\angle$$ DEA = $$\angle$$ AFD (Each 90°)
$$\angle$$ EAD = $$\angle$$ ADF (EA || DF)
$$\triangle$$ EAD $${\displaystyle \cong }$$ $$\triangle$$ 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),
$$BC^2 + AB^2 + AD^2 + DC^2 + 2EA × AB - 2AB × EA = DB^2 + AC^2$$
$$AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2$$