Question

If $ABCD$ is a rhombus, show that each diagonal bisects the vertex angle through which it passes.

Answer 1

Given that, $ABCD$ is a rhombus.

Hence, $AB=BC=CD=DA$

$In\Delta ABCand\Delta ADC$,

$AB=CD$

$BC=DA$

and

$\angle ABC=\angle ADC$ [Opposite angles of a rhombus are equal]

Hence, by SAS congruence criterion,

$\Delta ABC\cong \Delta ADC$

$\therefore \angle 1=\angle 4$ [cpct]

and $\angle 3=\angle 2$ [cpct]

Also,
$AB=BC$

$\therefore \Delta ABC$ is isosceles

Hence, $\angle 1=\angle 3$ [Angles opposite to equal sides of a triangle are equal]

$\therefore \angle 1=\angle 2=\angle 3=\angle 4$

Hence, it is proved that $AC$ bisects $\angle BCD$ and $\angle BAD$.

Similarly, $BD$ will bisect $\angle ABC$ and $\angle ADC$.