Question

Diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle A$ (see the given figure). Show that

(i) It bisects $\angle C$ also.

(ii) $ABCD$ is a rhombus.

Answer 1

Given:

In parallelogram $ABCD$,

$AC$ bisects $\angle A$.

To prove:

(i) It bisect $\angle C$ also

(ii) $ABCD$ is a rhombus.

Proof:

$ABCD$ is a parallelogram.

$\Rightarrow \angle DAC=\angle BCA$ (Alternate interior angles) ... (1)

And, $\angle BAC=\angle DCA$ (Alternate interior angles) ... (2)

However, it is given that $AC$ bisects $\angle A$.

$\Rightarrow \angle DAC=\angle BAC$ .... (3)

From equations (1), (2), and (3), we obtain

$\angle DAC=\angle BCA=\angle BAC=\angle DCA$ ... (4)

$\Rightarrow \angle DCA=\angle BCA$

Hence, $AC$ bisects $\angle C$ also.----(i)

From equation (4), we obtain

In $\Delta DAC$, $\angle DAC=\angle DCA$

$\Rightarrow DA=DC$ [Since, Sides opposite to equal angles are equal.]

However, $DA=BC$ and $AB=CD$ (Opposite sides of a parallelogram)

$\therefore AB=BC=CD=DA$

Therefore, $ABCD$ is a rhombus.---(ii)

Hence, proved both (i) and (ii).