Question

A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

Answer 1

Step 1: Drawing a diagram

Given: A diagonal of a parallelogram bisects one of its angles.

Let the parallelogram be $\mathrm{ABCD}$

Diagonal $\mathrm{AC}$ bisect $\angle \mathrm{A}$

Therefore, $\angle \mathrm{CAB}=\angle \mathrm{CAD}$

Step 2: Finding whether the parallelogram is rhombus or not

Here, $\mathrm{AB}\left|\right|\mathrm{CD}$ and $\mathrm{AC}$ is a transversal.

$\angle \mathrm{CAB}=\angle \mathrm{ACD}$ (Alternate interior angles)

Also, $\mathrm{AD}\left|\right|\mathrm{BC}$ and $\mathrm{AC}$ is a transversal.

$\angle \mathrm{DAC}=\angle \mathrm{ACB}$ (Alternate interior angles)

Thus, $\angle \mathrm{A}=\angle \mathrm{C}$ (As in a parallelogram opposite angles are equal)

$$\begin{gathered} \Rightarrow \frac{1}{2}\angle \mathrm{A}=\frac{1}{2}\angle \mathrm{C} \\ \Rightarrow \angle \mathrm{DAC}=\angle \mathrm{DCA} \\ \Rightarrow \mathrm{AD}=\mathrm{CD}(\mathrm{Sides}\mathrm{opposite}\mathrm{to}\mathrm{equal}\mathrm{angles}\mathrm{are}\mathrm{equal}) \end{gathered}$$

But, $\mathrm{AB}=\mathrm{CD}$ and $\mathrm{AD}=\mathrm{BC}$ (Opposite sides of parallelograms)

Therefore, $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}$

Thus, $\mathrm{ABCD}$ is a rhombus.