Chapter 4.2, Problem 2PSA

To determine

To Prove: $AC\perp BD.$

Explanation of Solution

Given information: ABCD is a rhombus.

Formula used: Properties of rhombus, linear pair axiom.

Proof:

  

  $\begin{array}{l}AB=BC=CD=DA\text{(given)}\\ \angle ADB=\angle DBC\&\angle DAC=\angle BCA(\text{Alternate linear angle})\\ \therefore \Delta AOD\cong \Delta COB(\text{ASA})\\ \therefore AO=CO\text{(CPCT)}\\ \text{OD=OD}\text{(Reflective property)}\\ \therefore \Delta AOD\cong \Delta COD(\text{SSS})\\ \therefore \Delta AOD=\Delta COD(\text{CPCT})\\ \angle AOD+\angle COD={180}^{\circ }(\text{Linear}\text{pair}\text{axiom})\\ \therefore \angle AOD={90}^{\circ }=\angle COD.\end{array}$