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Diagonal $ \mathrm{AC} $ of a parallelogram $ \mathrm{ABCD} $ bisects $ \angle \mathrm{A} $ (see below figure). Show that
(i) it bisects $ \angle \mathrm{C} $ also,
(ii) $ \mathrm{ABCD} $ is a rhombus.
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Academic Mathematics NCERT Class 9

Given:

Diagonal $( A C )$ of a parallelogram $( A B C D )$ bisects $\angle A$.

To do:

We have to show that:

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

(ii) $ABCD$ is a rhombus.

Solution:

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

$\therefore \angle DAC=\angle BAC$ ...... (1)

Now,

$AB\|DC$ and $AC$ as transversal,

$\therefore \angle BAC=\angle DCA$ [ Alternate angles ] ...... (2)

$AD\|BC$ and $AC$ as transversal,

$\therefore \angle DAC=\angle BCA$ [ Alternate angles ] .......(3)

From (1), (2) and (3)

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

$\therefore \angle DCA=\angle BCA$

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

(ii) In $\vartriangle ABC$,

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

$\Rightarrow BC=AB$ [ Sides opposite to equal angles are equal] ...... (1)

Also, $AB=CD$ and $AD=BC$ [Opposite sides of parallelogram are equal] .. (2)

From (1) and (2),

$\Rightarrow AB=BC=CD=DA$

Hence, $ABCD$ is a rhombus.

Updated on: 10-Oct-2022

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