Question

Diagonals of parallelogram ABCD intersect at O as shown in the figure. 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) $\angle OBY=\angle ODX$
(iii) $\angle BOY=\angle DOX$
(iv) $\Delta BOY\cong \Delta DOX$
Now, state if XY is bisected at O.

Answer 1

In parallelogram ABCD.

diagonals AC and BD intersect each other at O.
$\therefore$ O is the mid-point of AC and BD.
Throught O, XY is drawn such that X lies on AD and Y, on BC.
(i) OB = OD ($\because$ O is mid-point of BD)

(ii) $\because$ AD ||BC and BD is transversal
$\angle OBY=\angle ODX$ (Alternate angles)

(iii) $\angle BOY=\angle DOX$
(Vertically opposite angles)

(iv) Now in $\Delta BOY$ and $\Delta DOX$.
$\because OB=OD\angle OBY=\angle ODX\angle BOY=\angle DOX$
$\therefore \Delta BOY\cong \Delta DOX$ (ASA axiom)
$\therefore$ OY =OX (c.p.c.t)
Hence XY is bisected at O.