Question:
Diagonals of parallelogram ABCD intersect at O as shown in Fig. 17.30. XY contains O, and X, Y are points on opposite sides of the parallelogram. Give reasons for each of the following:
(i) OB = OD
(ii) ∠OBY = ∠ODX
(iii) ∠BOY = ∠DOX
(iv) ∆BOY ≅ ∆DOX
Now, state if XY is bisected at O..png)
Solution:
(i) Diagonals of a parallelogram bisect each other.
(ii) Alternate angles
(iii) Vertically opposite angles
(iv) In $\Delta$ BOY and $\Delta$ DOX:
$\mathrm{OB}=\mathrm{OD} \quad$ (diagonals of a parallelogram bisect each other)
$\angle \mathrm{OBY}=\angle \mathrm{ODX} \quad$ (a lternate a ngles)
$\angle \mathrm{BOY}=\angle \mathrm{DOX}$ (vertically o pposite a ngles)
ASA congruence:
XO = YO (c.p.c.t)
So, XY is bisected at O.