Chapter 9.5, Problem 12PSA

To determine

To prove: Diagonals of a square are congruent and perpendicular.

Explanation of Solution

Given information:Here, we have given that ABCD is a square with diagonals AC and BD.

Formula used: Distance formula: d = $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$ and Perimeter of quadrilateral = Sum of its all sides.

Proof:In $\Delta ACB$ and $\Delta BAD$ ,

  $BC\cong AD$ (Sides of square)

  $\angle ABC\cong \angle BAD$ (Right Angles)

  $AB\cong AB$ (Common)

  $\Rightarrow \Delta ACB\cong \Delta BAD$ (By SAS Congruency Rule)

  $\therefore AC\cong BD$ (By Corresponding Parts of Congruent Triangles)

Now, in $\Delta AOB$ and $\Delta AOD$ ,

  $OB\cong OD$ (Diagonals of square bisect each other)

  $AB\cong AD$ (Sides of Square)

  $OA\cong OA$ (Common)

  $\Rightarrow \Delta AOB\cong \Delta AOD$ (By SSS Congruency Rule)

  $\therefore \angle AOB\cong \angle AOD$ (By Corresponding Parts of Congruent Triangles)

As, $\angle AOB+\angle AOD=180°$ (Linear Pair angles)

  $\begin{array}{l}\Rightarrow \angle AOB+\angle AOB=180°\\ \Rightarrow 2(\angle AOB)=180°\\ \Rightarrow \angle AOB=\frac{180°}{2}\\ \Rightarrow \angle AOB=90°\end{array}$

And, $\angle AOD=90°$

  $\begin{array}{l}\Rightarrow AO\perp BD\\ \Rightarrow AC\perp BD\end{array}$