Question

$\square \mathrm{ABCD}$is a rhombus in which altitude from$\mathrm{D}$to side $\mathrm{AB}$ bisects $\mathrm{AB}$. Find the angles of the rhombus.

Answer 1

Finding the angles of the rhombus:

Given:

$\square \mathrm{ABCD}$ is a rhombus.

$\mathrm{DE}$ is the altitude on $AB$ then $\mathrm{AE}=\mathrm{EB}.$

From $\mathrm{\Delta AED}$ and $\mathrm{\Delta BED},$

Solution

Step1:

We note that,

$\mathrm{DE}=\mathrm{DE}$ (common line)

$\angle \mathrm{AED}=\angle \mathrm{BED}$(right angle, $\mathrm{DE}$ is an altitude)

$\mathrm{AE}=\mathrm{EB}$ (Given $\mathrm{DE}$ bisects $\mathrm{AB}$)

$\therefore \mathrm{\Delta AED}\cong \mathrm{\Delta BED}$by $\mathrm{SAS}$ property.

$\therefore \mathrm{AD}=\mathrm{BD}$ (by C.P.C.T)

Step2:

But $\mathrm{AD}=\mathrm{AB}$ (sides of rhombus are equal)

$\Rightarrow \mathrm{AD}=\mathrm{AB}=\mathrm{BD}$

$\therefore △\mathrm{ABD}$ is an equilateral triangle.

$\therefore \angle \mathrm{A}=60°$

Since, opposite angles of rhombus are equal, we get,

$\Rightarrow \angle \mathrm{A}=\angle \mathrm{C}=60°$

By the property of rhombus we know that sum of adjacent angles of a rhombus is supplementary.

$\begin{array}{rcl}\angle ABC+\angle BCD& =& 180°\\ \angle ABC+60°& =& 180°\\ \angle ABC& =& 180°–60°\\ & =& 120°\end{array}$

Since, opposite angles of rhombus are equal, we get,

$\angle \mathrm{ABC}=\angle \mathrm{ADC}=120°$

Hence, Angles of rhombus are:

$\angle A=60°,\angle C=60°,\angle B=120°,\angle D=120°$