Question

In a rhombus ABCD, the altitude from D to the side AB bisects AB. Find the angles of the rhombus.

Answer 1

Given: ABCD is a rhombus, DE is altitude which bisects AB i.e. AE = EB

$\mathrm{In}∆AED\mathrm{and}∆BED,$

$DE=DE$ (Common side)

$\angle DEA=\angle DEB=90°$ (Given)

$AE=EB$ (Given)

$\therefore ∆AED\cong ∆BED$ (By SAS congruence Criteria)

$\Rightarrow AD=BD$ (CPCT)

Also, $AD=AB$ (Sides of rhombus are equal)

$\Rightarrow AD=AB=BD$

Thus, $∆ABD$is an equilateral triangle.

Therefore, $\angle A=60°$

$\Rightarrow \angle C=\angle A=60°$ (Opposite angles of rhombus are equal)

$\mathrm{And},\angle ABC+\angle BCD=180°$ (Adjacent angles of rhombus are supplementary.)

$$\begin{gathered} \Rightarrow \angle ABC+60°=180° \\ \Rightarrow \angle ABC=180°-60° \\ \Rightarrow \angle ABC=120° \\ \Rightarrow \angle ADC=\angle ABC=120° \end{gathered}$$

Hence, the angles of rhombus are $60°,120°,60°\mathrm{and}120°$.