Diagonals of parallelogram ABCD intersect at O as shown in Fig. 17.30. XY contains O,


Diagonals of parallelogram ABCD intersect at O as shown in Fig. 17.30. XY contains O, and XY 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.


(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.

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