# 1. In Fig. 6.28, find the values of x and y and then show that AB $||$ CD.

Given that,
In the figure, CD and PQ intersect at  F
Therefore, $y =130^0$  (vertically opposite angles)

PQ is a straight line. So,
$\\x + 50^0 =180^0\\ x = 130^0$

Hence AB || CD (since $x$ and are $y$ alternate interior angles)

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