Chapter 4.3, Problem 3PSA

To determine

To Prove: Diagonals of rhombus are perpendicular to each other.

Explanation of Solution

Given information: ABCD is a Rhombus.

Formula used: Properties of rhombus, Congruency of triangles, linear pair axiom.

Proof:

  

  $\begin{array}{l}ABCD\text{is}\text{a}\text{rhombus}\text{(Given)}\\ \therefore AB=BC=CD=DA.\\ BD=BD(\text{Reflection}\text{property})\\ \therefore \Delta BAD\cong \Delta BCD(\text{SSS})\\ \therefore \angle ABE=\angle CBE(CPCT)\\ BE=BE(\text{Reflection}\text{property})\\ \therefore \Delta ABE\cong \Delta CBE(\text{SAS})\\ \therefore \angle AEB=\angle CEB(\text{CPCT})\\ \angle AEB+\angle CEB={180}^{\circ }(\text{linear}\text{pair}\text{axiom})\\ \therefore \angle AEB={90}^{\circ }=\angle CEB.\\ \therefore AC\perp BD\end{array}$