If diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angle. Also, prove that it is a rhombus.

Explanation

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

`therefore" "angleCAB = angleCAD" "...(i)`
To show ABCD is a rhombus.
Proof Since, ABCD is a parallelogram, therefore `AB||CD and AC` is a transversal.
`therefore" "angleCAB=angleACD" "`[alternate interior angles]
Again, `AD||BC and AC` is a transversal.
`therefore" "angleCAD=angleACB" "`[alternate interior angles]
So, `" "angleACD=angleACB" "[becauseangleCAB=angleCAD,given]...(ii)`
Also, `" "angleA=angleC` [opposite angles of parallelogram are equal]
`rArr" "(1)/(2)angleA=(1)/(2)angleC" "` [dividing both sides by 2]
`rArr" "angleDAC=angleDCA" "`[from Eqs. (i) and (ii)]
`rArr" "CD=AD`
`" "` [sides opposite to the equal angles are equal]
But `" "AB=CD and AD=BC`
`" "` [opposite sides of parallelogram are equal]
`therefore" "AB=BC=CD=AD`
Thus, all sides are equal. So, ABCD is a rhombus. `" "` Hence proved.