Question

In case of a parallelogram
prove that:
(i) The bisectors of any two adjacent angles intersect at 90^o.
(ii) The bisectors of the opposite angles are parallel to each other.

Answer 1

ABCD is a parallelogram, the bisectors of ∠ADC and ∠BCD meet at a point E and the bisectors of ∠BCD and ∠ABC meet at F.

We have to prove that the ∠CED = 90° and ∠CFG = 90°

Proof: In the parallelogram ABCD
∠ADC + ∠BCD = 180°       ....[ sum of adjacent angles of a parallelogram ]

⇒ `"∠ADC"/2 + "∠BCD"/2` = 90°

⇒ ∠EDC + ∠ECD + ∠CED = 180°
⇒ ∠CED = 90°

Similarly taking triangle BCF it can be proved that ∠BFC = 90°
∠BFC + ∠CFG = 180°               ....[ adjacent angles on a line ]
Also ⇒ ∠CFG = 90°
Now since ∠CFG = ∠CED = 90° ....[ It means that the lines DE and BG are parallel ]
Hence proved.