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Answer :

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)

AD = AD (Common Angles)

\(\triangle\) 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),

\(BC^2 + AB^2 + AD^2 + DC^2 + 2EA × AB - 2AB × EA = DB^2 + AC^2\)

\(AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2\)

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