Ex. 6.6 Q6 Triangles Solution - NCERT Maths Class 10

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

Text Solution

Reasoning:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Steps:

In parallelogram \(ABCD \)

\[\begin{align} AB = CD \\ AD = BC\end{align}\]

Draw \(A E \perp C D,\) \(D F \perp A B\) 

\[\begin{align} ​​​​& \qquad \qquad EA = DF \\& \begin{bmatrix} \text{Perpendiculars drawn between }\\ \text{same parallel lines}\end{bmatrix}  \end{align}\]

In \(\Delta AEC\)

\begin{align}A{C^2} &\! = \! A{E^2} + E{C^2}\\ &\! =\!  A{E^2} + {\left[ {ED + DC} \right]^2}\\ &\! = \! A{E^2} \! \!+\!\! D{E^2} \!\!+\! \! D{C^2} \!+\! 2DE.\! DC \\ A{C^2} &\! = \! A{D^2}\! + \! D{C^2} \! + \! 2DE \cdot \! DC  \; \ \ldots \left(\text{i} \right) \\ & \left[ {{\text{Since, }}A{D^2} \! =\!  A{E^2} + D{E^2}} \right]\end{align}

In \(\Delta \,DFB\)

\[\begin{align} B{D^2} &\!=\! D{F^2}\! +\! B{F^2}\\ &\!=\! D{F^2}\! +\! {\left[ {AB\! -\! AF} \right]^2}\\ &\!= \! D{F^2} \!\! +\!\! A{B^2} \!\! +\!\! A{F^2}\! \!-\!\! 2AB. \!AF \\ &\!=\! A{D^2} \!+ \!A{B^2}\! - \!2AB.AF\\ B{D^2} &\!= \!A{D^2} \!+\! A{B^2}\! -\! 2AB.AF \; \ldots \left( \rm{ii} \right) \\ &\text{(Since } A{D^2} = D{F^2} + A{F^2}) \end{align}\]

Adding (i) and (ii)

\[\begin{align} A{C^2}\!+\!B{D^2} &\!=\! \begin{bmatrix}A{D^2} \!+\! D{C^2} \! +\! 2DE \cdot DC \!+\! \\ A{D^2} \!+\! A{B^2} \! -\! 2AB.AF \end{bmatrix} \\ A{C^2} \!+\! B{D^2} &\!= \!\begin{bmatrix}B{C^2}\! +\! D{C^2} \!+\! A{D^2} \!+\! A{B^2} \!+\!\\ 2AB.AF \!- \!2AB.F\end{bmatrix} \\  \text{(Since } &AD \!=\! BC, \;DE \!=\! AF,\; CD \!=\! AB) \end{align}\]

\[\begin{align} \Rightarrow \quad& A{C^2} + B{D^2} \\  = & A{B^2} \!+\! B{C^2} \!+\! C{D^2} \!+ \! A{D^2} \end{align} \]