Chapter 5.5, Problem 22PSB

To determine

To prove: one diagonal of a kite divides it into two congruent triangles, while the other diagonal divides it into two isosceles triangles.

Explanation of Solution

Given information:

  $ABCD$ is a kite, $AB=AD,BC=CD$ and $AC\text{ \& }BD$ are diagonals.

Formula used:

SSS Rule: If the corresponding sides of two triangles are equal in lengths, then two triangles are congruent.

Proof:

From the figure, the diagonal $AC$ divides the kite $ABCD$ into two triangles, $\Delta ACD\text{ and }\Delta \text{ACB}$

  

  $\begin{array}{l}\text{In }\Delta ACD\text{ and }\Delta ACB\\ AC=AC\left[\because \text{ Common sides}\right]\\ AD=AB\left[\because \text{ From given}\right]\\ CD=CB\left[\because \text{ From given}\right]\\ \Rightarrow \Delta ACD\cong \Delta ACB\end{array}$

This implies that $AC$ divides the kite $ABCD$ into two congruent triangles, $\Delta ACD\cong \Delta ACB$

From the figure, the diagonal $BD$ divides the kite $ABCD$ into two triangles, $\Delta ADB\text{ and }\Delta \text{CDB}$

  

  $\begin{array}{l}\text{In }\Delta ADB\text{ and }\Delta \text{CDB}\\ DB=DB\left[\because \text{ Common sides}\right]\\ AD=AB\left[\because \text{ From given}\right]\\ CD=CB\left[\because \text{ From given}\right]\end{array}$

  $AD=AB$ implies $\Delta ADB$ is an isosceles triangle and

  $CD=CB$ implies $\Delta \text{CDB}$ is an isosceles triangle

This implies that $BD$ divides the kite $ABCD$ into two isosceles triangles, $\Delta ADB\text{ and }\Delta \text{CDB}$ .