Diagonal AC of a paraleligram ABCD bisects `angleA` (sec figure). Show that:
(i) it bisects `angleC` also (ii) ABCD is a rhombus.

Explanation

Given, diagonal AC of a parallelogram ABCD bisects `angleA.`
(i) We have


`angle1=angle2" "("given")...(1)`
`But" "angle1=angle4("alternate angles as" AB"||"DC)...(2)`
`and" "angle2=angle3" "("alternate angles as"AD"||"BC)...(3)`
`therefore` Diagonal AC also bisects `angleC.`
(ii) `"Since"" "angle1=angle2" "("given")`
`"and"" "angle2=angle3" "("alternate angles as" "||"BC)`
`therefore" "angle1=angle3`
`implies" "AB=BC" "("sides opposite to equal angles are equal")`
Now, since djacent sides of a parallelogram are equal.
`therefore" "squareABCD` is a rhombus.