Question

Prove that sum of square of diagonal of parallelogram is equal to sum of squares of its sides

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Sum of square of diagonal of parallelogram is equal to sum of squares of its sides is proved

Solution:

The diagram for this sum is attached below

In parallelogram ABCD, AB = CD, BC = AD

Draw perpendiculars from C and D on AB as shown.  

In right angled Δ AEC,

$\begin{array}{l}{\mathrm{AC}^{2}=\mathrm{AE}^{2}+\mathrm{CE}^{2}[\mathrm{By} \text { Pythagoras theorem }]} \\\\ {\Rightarrow \mathrm{AC}^{2}=(\mathrm{AB}+\mathrm{BE})^{2}+\mathrm{CE}^{2}} \\\\ {\Rightarrow \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BE}^{2}+2 \mathrm{AB} \times \mathrm{BE}+\mathrm{CE}^{2} \rightarrow(1)}\end{array}$

From the figure CD = EF (Since CDFE is a rectangle)  

But CD= AB  

⇒ AB = CD = EF  

Also CE = DF (Distance between two parallel lines)

ΔAFD ≅ ΔBEC (RHS congruence rule)  

⇒ AF = BE  

Consider right angled ΔDFB

$\begin{array}{l}{\mathrm{BD}^{2}=\mathrm{BF}^{2}+\mathrm{DF}^{2}[\mathrm{By} \text { Pythagoras theorem }]} \\\\ {=(\mathrm{EF}-\mathrm{BE})^{2}+\mathrm{CE}^{2}[\text { since DF }=\mathrm{CE}]} \\\\ {=(\mathrm{AB}-\mathrm{BE})^{2}+\mathrm{CE}^{2}[\text { Since } \mathrm{EF}=\mathrm{AB}]} \\\\ {\Rightarrow \mathrm{BD}^{2}=\mathrm{AB}^{2}+\mathrm{BE}^{2}-2 \mathrm{AB} \times \mathrm{BE}+\mathrm{CE}^{2} \rightarrow(2)}\end{array}$

Add (1) and (2),

$\mathrm{AC}^{2}+\mathrm{BD}^{2}=\left(\mathrm{AB}^{2}+\mathrm{BE}^{2}+2 \mathrm{AB} \times \mathrm{BE}+\mathrm{CE}^{2}\right)+\left(\mathrm{AB}^{2}+\mathrm{BE}^{2}-2 \mathrm{AB} \times \mathrm{BE}+\mathrm{CE}^{2}\right)$

$\begin{array}{l}{=2 \mathrm{AB}^{2}+2 \mathrm{BE}^{2}+2 \mathrm{CE}^{2}} \\\\ {\mathrm{AC}^{2}+\mathrm{BD}^{2}=2 \mathrm{AB}^{2}+2\left(\mathrm{BE}^{2}+\mathrm{CE}^{2}\right) \rightarrow(3)}\end{array}$

$\begin{array}{l}{\text { From right angled } \Delta \mathrm{BEC}, \mathrm{BC}^{2}=\mathrm{BE}^{2}+\mathrm{CE}^{2}[\mathrm{By}\text { Pythagoras theorem] } } \\\\ {\text { Hence equation }(3) \text { becomes, } \mathrm{AC}^{2}+\mathrm{BD}^{2}=2 \mathrm{AB}^{2}+2 \mathrm{BC}^{2}} \\\\ {=\mathrm{AB}^{2}+\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{BC}^{2}} \\\\{=\mathrm{AB}^{2}+\mathrm{CD}^{2}+\mathrm{BC}^{2}+\mathrm{AD}^{2}}\end{array}$

$\mathrm{AC}^{2}+\mathrm{BD}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CD}^{2}+\mathrm{AD}^{2}$

Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Learn more about parallelogram

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

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