Question

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

Answer 1

Given Let ABCD is a parallelogram and diagonal AC bisects the angle A.

$\therefore \angle CAB=\angle CAD$ . . . . . . . (i)

To show ABCD is a rhombus.

Proof Since, ABCD is a parallelogram, therefore AB || CD and AC is a transversal.

$\therefore \angle CAB=\angle ACD$ [alternate interior angles]

Again, AD || BC and AC is a transversal.

$\therefore \angle CAD=\angle ACB$ [alternate interior angles]

So, $\angle ACD=\angle ACB$ [$\because \angle CAB=\angle CAD$ given] . . . . (ii)

Also, $\angle A=\angle C$ [opposite angles of parallelogram are equal]

$\Rightarrow \frac{1}{2}\angle A=\frac{1}{2}\angle C$ [dividing both sides by 2]

$\Rightarrow \angle DAC=\angle DCA$ [from Eqs. (i) and (ii)]

$\Rightarrow CD=AD$

[sides opposite to the equal angles are equal]

But AB = CD and AD = BC

[opposite sides of parallelogram are equal]

$\because AB=BC=CD=AD$

Thus, all sides are equal. So, ABCD is a rhombus. Hence proved.