If a diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

Name: If a diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.
Uploaded: Thu, 2021-01-07 23:41
Description: If a diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

If a diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

Answers (1)

Let ABCD be a parallelogram and diagonal AC bisect the angle A and $\angle CAB = \angle CAD$

To Prove: ABCD is a rhombus
Proof: Here it is given that ABCD is a parallelogram.
$AB\parallel CD$ and AC is transversal
$\angle CAB=\angle ACD$ {alternate interior angles} …. (1)
Also $AD\parallel BC$ an AC is a transversal
$\angle CAD=\angle ACB$ {alternate interior angle} …. (2)
Now it is given that $\angle CAB = \angle CAD$
So from equations (1) and (2)
$\Rightarrow \angle ACD=\angle ACB$ …. (3)
Also, $\angle A=\angle C$ {opposite angles of a parallelogram}
$\frac{\angle A}{2}=\frac{\angle C}{2}$ {dividing by 2}
$\angle DAC=\angle DCA$ {from 1 and 2}
$CD=AD$ {sides opposite of equal angles in $\triangle ADC$ are equal}
$AB=CD$ & $AD=BC$ { $\because$ opposite sides of a parallelogram}
Hence,
$AB = BC = CD = AD$
Thus all sides are equal so ABCD is a rhombus.
Hence Proved.