Question

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

Answer 1

Given: ABCD is a parallelogram
In the parallelogram,
AB = DC
AD = BC

Step - 1: Constructution: Draw $\text{AF}\perp \text{DC}$, extend A to E and $\text{ED}\perp \text{EB}$
Draw perpendiculars DE and AF

Step - 2: Applying A-A similarity in $\Delta \text{ADF and}\Delta \text{DAE}$
$\angle \text{AFD}=\angle \text{AED}={90}^{\circ }$ [By construction]
$\angle \text{ADF}=\angle \text{EAD}$ [alternae interior angles as lines ES || DC cut by a transversal AD]
$\therefore \Delta \text{ADF~}\Delta \text{DAE}$ [By A-A similarity criteria] ...(1)
AE = DF [CPST]

Step 3: Applying Pythagoras theorem in $\Delta \text{DAE,}\Delta \text{DEB,}\Delta \text{ADF,}\Delta \text{AFC,}$
Applying Pythagoras theorem in $\Delta \text{DAE,}$
On applying Pythagoras theorem,
${\text{DA}}^2={\text{DE}}^2+{\text{EA}}^2$ ...(2)

In $\Delta \text{DEB}$
On applying Pythagoras theorem,
${\text{DB}}^2={\text{DE}}^2+{\text{EB}}^2$
${\text{DB}}^2={\text{DE}}^2+(\text{EA+AB}{)}^2$
${\text{DB}}^2={\text{DE}}^2+{\text{EA}}^2+{\text{AB}}^2+\text{2.EA.AB}$ ...(3)
$\text{In}\Delta \text{ADF}$
On applying Pythagoras theorem,
${\text{DA}}^2={\text{DF}}^2+{\text{AF}}^2$ ...(4)

In $\Delta \text{AFC}$
On applying Pythagoras theorem,

${\text{AC}}^2={\text{AF}}^2+{\text{FC}}^2$
${\text{AC}}^2={\text{AF}}^2+(\text{DC-FD}{)}^2$
${\text{AC}}^2={\text{DA}}^2+{\text{DC}}^2-2.\text{DC.FD}$ ...(5)

Step - 4: Adding equation (3) and (5)
${\text{DB}}^2+{\text{AC}}^2={\text{DA}}^2+{\text{DC}}^2-\text{2.DC.FD}+{\text{DA}}^2+{\text{AB}}^2+\text{2.EA.AB}$ ...(6)
In the parallelogram,
AB = DC, AD = BC ...(given)
$\Delta \text{ADF~}\Delta \text{DAE}$ [from 1]
$\therefore$ EA = DF [CPST]

Substitute the above values in equation (6)
$\Rightarrow {\text{DB}}^2+{\text{AC}}^2={\text{BC}}^2+{\text{DC}}^2-\text{2.AB.EA}+{\text{DA}}^2+{\text{AB}}^2+\text{2.EA.AB}$
$\Rightarrow {\text{DB}}^2+{\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2+{\text{DC}}^2+{\text{DA}}^2$
$\Rightarrow {\text{DB}}^2+{\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2+{\text{DC}}^2+{\text{DA}}^2$
Sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Hence proved.