Chapter 9.5, Problem 25PSC

To determine

To prove: In quadrilateral ABCD , the sum of the square of the sides AB , BC , CD and DA is equal to the sum of the square of the diagonals AC and BD plus four times the square of the segment MN joining the midpoints of the diagonals AC and BD .

Explanation of Solution

Given information:

In quadrilateral ABCD, AB , BC , CD and DA are sides.

AC and BD are diagonals. MN is the segment joining the midpoints of the diagonals AC and BD .

Formula used:

The distance between two points by using the distance formula, which is an application of the Pythagoras theorem.

D $=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Proof:

  

The distance AB , BC, CD , DA , AC , BD and MN can be calculated by applying distance formula.

Distance AB :

  $\begin{array}{l}{\left(AB\right)}^2={\left(a-0\right)}^2+{\left(0-0\right)}^2\\ {\left(AB\right)}^2=a^2\end{array}$

Distance BC :

  $\begin{array}{l}{\left(BC\right)}^2={\left(a-d\right)}^2+{\left(e-0\right)}^2\\ {\left(BC\right)}^2=a^2-2ad+d^2+e^2\end{array}$

Distance CD :

  $\begin{array}{l}{\left(CD\right)}^2={\left(b-d\right)}^2+{\left(c-e\right)}^2\\ {\left(CD\right)}^2=b^2-2bd+d^2+c^2-2ce+e^2\\ {\left(CD\right)}^2=a^2-2ab+b^2+c^2\end{array}$

Distance DA :

  $\begin{array}{l}{\left(DA\right)}^2={\left(b-0\right)}^2+{\left(c-0\right)}^2\\ {\left(DA\right)}^2=b^2+c^2\end{array}$

Distance AC :

  $\begin{array}{l}{\left(AC\right)}^2={\left(d-0\right)}^2+{\left(e-0\right)}^2\\ {\left(AC\right)}^2=d^2+e^2\end{array}$

Distance BD :

  $\begin{array}{l}{\left(BD\right)}^2={\left(a-b\right)}^2+{\left(0-c\right)}^2\\ {\left(BD\right)}^2=a^2-2ab+b^2+c^2\end{array}$

Distance MN :

  $\begin{array}{l}{\left(MN\right)}^2={\left(\frac{a+b}{2}-\frac{d}{2}\right)}^2+{\left(\frac{c}{2}-\frac{e}{2}\right)}^2\\ {\left(MN\right)}^2=\frac{a^2+b^2+d^2+2ab-2ad-2bd}{4}+\frac{c^2-2ce+e^2}{4}\\ 4{\left(MN\right)}^2=a^2+b^2+d^2+2ab-2ad-2bd+c^2-2ce+e^2\end{array}$

Adding squares of all sides of quadrilateral ABCD , we get

  ${\left(AB\right)}^2+{\left(BC\right)}^2+{\left(CD\right)}^2+{\left(DA\right)}^2=2a^2+2b^2+2c^2+2d^2+2e^2-2ad-2bd-2ce$

Adding sum of the square of the diagonals AC and BD plus four times the square of the segment MN joining the midpoints of the diagonals AC and BD , we get

  ${\left(AC\right)}^2+{\left(BD\right)}^2+4{\left(MN\right)}^2=2a^2+2b^2+2c^2+2d^2+2e^2-2ad-2bd-2ce$

  $\therefore {\left(AB\right)}^2+{\left(BC\right)}^2+{\left(CD\right)}^2+{\left(DA\right)}^2={\left(AC\right)}^2+{\left(BD\right)}^2+4{\left(MN\right)}^2$