Question

6. In the figure 10.12, if ZBAO = ZDCO and OC = OD, show that AB || CD.
Fig. 10.12​

Answer 1

As per given figure, following statements are given as true:

1. $\angle BAO = \angle DCO$

2. Side OC = Side OD

To prove: Side AB || CD

Solution:

It is given that $\angle BAO = \angle DCO$

Let $\angle BAO = \angle DCO =x$ ....... (1)

Now, let us consider $\triangle OCD$:

Given that Side OC = OD

i.e. $\triangle OCD$ is an isosceles triangle.

Angles opposite to equal sides are also equal.

$\Rightarrow \angle DCO = \angle CDO = x\ (\text{As per equation (1)})$

$\therefore \angle BAO = \angle CDO = x$

$\angle BAO\ and\ \angle CDO$ are the two angles formed by intersection of lines AB and CD by the common line AD.

$\therefore$ the two angles $\angle BAO\ and\ \angle CDO$ are alternate angels which is possible only when the sides AB and CD are parallel to each other.

Hence proved that AB || CD.