Question

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

Answer 1

To prove: (i) $AC$ bisects $\angle C$

(ii) $ABCD$ is a rhombus.

(i) Here, $ABCD$ is a parallelogram and diagonal $AC$ bisects $\angle A.$

$\therefore \angle DAC=\angle BAC\dots \dots \left(i\right)$

Now, $AB\parallel DC$ and $AC$ is the traversal, we have

$\therefore \angle BAC=\angle DCA\dots \dots \left(ii\right)$ [ Alternate interior angles]

$AD\parallel BC$ and $AC$ as traversal,

$\therefore \angle DAC=\angle BCA\dots \dots \left(iii\right)$ [ Alternate interior angles ]

From eq. (i), (ii), and (iii), we have

$\angle DAC=\angle BAC=\angle DCA=\angle BCA$

$\therefore \angle DCA=\angle BCA$

Hence, $AC$ bisects $\angle C.$

(ii) In $△ABC,$ we have

$\Rightarrow \angle BAC=\angle BCA$ [ Proved in above ]

$\Rightarrow BC=AB\dots \dots \left(iv\right)$ [ Sides opposite to equal angles are equal ]

$\Rightarrow$ Also, $AB=CD$ and $AD=BC\dots \dots \left(v\right)$ [ Opposite sides of parallelogram are equal ]

From eq.(iv) and (v), we have

$\Rightarrow AB=BC=CD=DA$

Hence, $ABCD$ is a rhombus.