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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.

Q6   Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

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In parallelogram ABCD, AF and DE are altitudes drawn on DC and produced BA.

In $\triangle$DEA, by Pythagoras theorem

$DA^2=DE^2+EA^2.......................1$

In $\triangle$DEB, by Pythagoras theorem

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

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

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

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

In $\triangle$ADF, by Pythagoras theorem

$DA^2=AF^2+FD^2$

In $\triangle$AFC, by Pythagoras theorem

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

$\Rightarrow AC^2=AF^2+(DC)^2+(FD)^2-2.DC.FD$

$\Rightarrow AC^2=(AF^2+FD^2)+(DC)^2-2.DC.FD$

$\Rightarrow AC^2=AD^2+(DC)^2-2.DC.FD.......................3$

Since ABCD is a parallelogram.

SO, AB=CD  and BC=AD

In $\triangle DEA\, and\, \triangle ADF,$

$\angle DEA=\angle AFD\, \, \, \, \, \, \, (each 90 \degree)$

$\angle DAE=\angle ADF$      (AE||DF)

$\triangle DEA\, \cong \, \triangle ADF,$        (ASA rule)

$\Rightarrow EA=DF.......................6$

Adding 2 and, we get

$DA^2+AB^2+2.EA.AB+AD^2+DC^2-2.DC.FD=DB^2+AC^2$$\Rightarrow DA^2+AB^2+AD^2+DC^2+2.EA.AB-2.DC.FD=DB^2+AC^2$

$\Rightarrow BC^2+AB^2+AD^2+2.EA.AB-2.AB.EA=DB^2+AC^2$   (From 4 and 6)

$\Rightarrow BC^2+AB^2+CD^2=DB^2+AC^2$

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