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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)

AD=AD (common)

$\triangle DEA\, \cong \, \triangle ADF,$ (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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