Question

Diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle A$. Show that
(i) It bisects $\angle C$ also,
(ii) $ABCD$ is a rhombus.

Answer 1

$\square ABCD$ is a parallelogram ....given

side $AD\cong$ side $BC$ ...opposite sides of a parallelogram are congruent ....(1)

In $△ADC$ and $△ABC$,

side $AD\cong$ side $BC$ ....from (1)

$\angle DAC=\angle ACB$ .....Alternate angle

side $AC\cong$ side $AC$ ....common side

$△ADC\cong △ABC$ .....SAS test of congruence

$\therefore \angle DCA=\angle BCA$ ....c.a.c.t. ....(2)

$\therefore$ Diag $AC$ is bisector of $\angle C$.

We know, opposite angles of a parallelogram are congruent.

$\angle DAB=\angle BCD$

$\therefore \frac{1}{2}\angle DAB=\frac{1}{2}\angle BCD$

$\therefore \angle BAC=\angle BCA$ ....(3)

In $△ABC$,

$\angle BAC=\angle BCA$ ....from (3)

$\therefore BC=AB$ ....Converse of isosceles triangle theorem

Similarly, we can prove $AD=DC$.

Since, the adjacent sides of a parallelogram are equal,

$\square ABCD$ is a rhombus.