Question

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

Answer 1

Given that $ABCD$ is a rhombus and the altitude on $AB$ such that $AE=EB.$

In a $△AED$ and $△BED,$ we have

$DE=DE$ ( common line)

$\angle AED=\angle BED$ ( right angle)

$AE=EB$ ( DE bisects AB)

$\therefore △AED\cong △BED$ ( by SAS property)

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

But $AD=AB$ ( sides of rhombus are equal)

$\Rightarrow AD=AB=BD$

$\therefore$ ABD is an equilateral triangle.

$\therefore \angle A={60}^{\circ }$

$\Rightarrow \angle A=\angle C={60}^{\circ }$ (opposite angles of rhombus are equal)

But Sum of adjacent angles of a rhombus is supplementary.

$\angle ABC+\angle BCD={180}^{\circ }$

$\Rightarrow \angle ABC+{60}^{\circ }={180}^{\circ }$

$\Rightarrow \angle ABC={180}^{\circ }-{60}^{\circ }={120}^{\circ }$

$\therefore \angle ABC=\angle ADC={120}^{\circ }$ (opposite angles of rhombus are equal)

Hence, Angles of rhombus are $\angle A={60}^{\circ }$ and $\angle C={60}^{\circ }$, $\angle B=\angle D={120}^{\circ }.$