Chapter 9.5, Problem 24PSB

To determine

To prove:The sum of square of the sides and diagonals of a parallelogram is equal.

Explanation of Solution

Given information:Here, we have given a parallelogram ABCD.

Formula used: Pythagoras Theorem: ${(H)}^2={(P)}^2+{(B)}^2$ , ${(a+b)}^2=a^2+b^2+2ab$

Proof: Draw perpendiculars from C and D on AB.

Now, In $\Delta AEC$ ,

  ${(AC)}^2={(AE)}^2+{(CE)}^2$ (By Pythagoras theorem)

  $\Rightarrow A{C}^2={(AB+BE)}^2+{(CE)}^2$

  $\Rightarrow A{C}^2 =A{B}^2 +B{E}^2 +2\text{ (}AB)(BE)+C{E}^2$ (Equation 1)

From the figure, $CD=EF$ . (CDEF is a rectangle)

  $CD=AB$ (Opposite sides of parallelogram are equal)

  $\Rightarrow AB=EF$

Now, In $\Delta AFD$ and $\Delta BEC$ ,

  $\angle AFD\cong \angle BEC$ (Right angles)

  $CE\cong DF$ (Distance between parallel lines is equal)

  $CD\cong EF$ (Proved above)

  $\Rightarrow \Delta AFD\cong \Delta BEC$ (By RHS Congruency Rule)

  $\therefore AF=BC$ (By Corresponding Parts of Congruent triangles)

Now, In $\Delta DFB$ ,

  $B{D}^2=B{F}^2 +D{F}^2$ (By Pythagoras Theorem)

  $\Rightarrow B{D}^2={\left(EF–BE\right)}^2+C{E}^2$ (As, $BF=EF-BE$ and $CE=DF$ )

  $\Rightarrow B{D}^2={\left(AB–BE\right)}^2 +C{E}^2$ (As, $AB=EF$ )

  $\Rightarrow B{D}^2 =A{B}^2 +B{E}^{} –2(AB)(BE)+C{E}^2$ (Equation 2)

Now, Adding Equation 1 and Equation 2, we get

  $A{C}^2 +B{D}^2=\text{[}A{B}^2 +B{E}^2 +2(AB)(BE)+C{E}^2]+[A{B}^2 +B{E}^2 –2(AB\text{)(}BE)+C{E}^2]$

  $A{C}^2 +B{D}^2 =2A{B}^2 +2B{E}^2 +2C{E}^2$

  $\Rightarrow A{C}^2 +B{D}^2=2A{B}^2 +2(B{E}^2 +C{E}^2)$ (Equation 3)

Now, In $\Delta BEC$ ,

  $B{C}^2=B{E}^2 +{\text{ CE}}^2$ (By Pythagoras theorem)

  $\Rightarrow A{C}^2 +B{D}^2=2A{B}^2 +2B{C}^2$ (From Equation 3)

  $\Rightarrow A{C}^2 +B{D}^2=A{B}^2 +A{B}^2+B{C}^2 +B{C}^2$

  $\Rightarrow A{C}^2 +B{D}^2= A{B}^2 +C{D}^2 +B{C}^2 +A{D}^2$ (Opposite sides of parallelogram are equal)

  $\therefore A{C}^2+B{D}^2=A{B}^2+B{C}^2+C{D}^2+A{D}^2$

  $\therefore$ The sum of square of the sides and diagonals of a parallelogram is equal.