Chapter 5.5, Problem 18PSB

To determine

To prove: the diagonals of a Rhombus divide the rhombus into four congruent triangles.

Explanation of Solution

Given information:

• Let ABCD be a Rhombus and AC & BD be diagonals of the Rhombus.
• Diagonals of Rhombus ABCDS bisect each other.
• Two triangles are congruent if the corresponding sides are equal in length (from SSS Rule).

Formula used:

All sides of a Rhombus are equal in size.

Proof:

  

Given ABCD is a Rhombus, so all the sides are equal in size, i.e, $AB=BC=CD=DA$

  $\begin{array}{l}\text{From }\Delta AOB\text{ \& }\Delta AOD\\ AB=AD\left[\because \text{ Sides are equal}\right]\\ AO=AO\left[\because \text{ Common Sides}\right]\\ OD=OB\left[\because \text{ Diagonals bisect each other}\right]\\ \text{Implies that }\Delta AOB\cong \Delta AOD\dots \left(i\right)\end{array}$

  

  $\begin{array}{l}\text{From }\Delta AOB\text{ \& }\Delta BOC\\ AB=BC\left[\because \text{ Sides are equal}\right]\\ BO=BO\left[\because \text{ Common Sides}\right]\\ AO=OC\left[\because \text{ Diagonals bisect each other}\right]\\ \text{Implies that }\Delta AOB\cong \Delta BOC\dots \left(ii\right)\end{array}$

  

Similarly,

  $\begin{array}{l}\text{From }\Delta BOC\text{ \& }\Delta COD\\ BC=CD\left[\because \text{ Sides are equal}\right]\\ CO=CO\left[\because \text{ Common Sides}\right]\\ BO=OD\left[\because \text{ Diagonals bisect each other}\right]\\ \text{Implies that }\Delta BOC\cong \Delta COD\dots \left(iii\right)\end{array}$