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.

Answer 1

(i) Diagonals of a parallelogram bisect each other.
(ii) Alternate angles
(iii) Vertically opposite angles
(iv)
$$\begin{gathered} \mathrm{In}∆\mathrm{BOY}\mathrm{and}∆\mathrm{DOX}: \\ \mathrm{OB}=\mathrm{OD}(d\mathrm{iagonals}\mathrm{of}\mathrm{a}\mathrm{parallelogram}\mathrm{bisect}\mathrm{each}\mathrm{o}\mathrm{ther}) \\ \angle \mathrm{OBY}=\angle \mathrm{ODX}(a\mathrm{lternate}\mathrm{a}\mathrm{ngles}) \\ \angle \mathrm{BOY}=\angle \mathrm{DOX}(v\mathrm{ertically}\mathrm{o}\mathrm{pposite}\mathrm{a}\mathrm{ngles}) \end{gathered}$$


ASA congruence:
XO = YO (c.p.c.t)
So, XY is bisected at O.