91Ó°ÊÓ

The given diagram is:

It is being given that Plane $P$$âˆ\yen$ Plane $Q$ and $jâˆ\yen k$.

As the points $A$ and $B$ are lying on the plane $P$, therefore the line $AB$ joining the points $A$ and $B$ is also lying on the plane $P$.

As the points $X$ and $Y$ are lying on the plane $Q$, therefore the line $XY$ joining the points $X$ and $Y$ is also lying on the plane $Q$.

As the planes $P$ and $Q\stackrel{¯}{\mathbf{AC}}$ are parallel, therefore the lines lying on the planes are also parallel.

Therefore, the lines $AB$ and $XY$ are parallel.

That implies,$ABâˆ\yen XY$.

As, $jâˆ\yen k$, therefore the segments of the lines $j$ and $\stackrel{¯}{AC}k$are also parallel.

Therefore, $AXâˆ\yen BY$.

In the parallelogram pair of opposite sides are parallel and congruent.

Therefore, as $ABâˆ\yen XY$ and $AXâˆ\yen BY$ that implies the pair of opposite sides are parallel.

Therefore, $ABYX$ is a parallelogram.

Therefore, the opposite sides of the parallelogram are congruent.

Therefore, $AXâ‰\dots BY$.

Therefore, $\stackrel{¯}{BD}AX=BY$.

Hence proved.

The given diagram is:

The theorem 3-1 states that if two parallel planes are cut by a third plane, then the lines of intersection are parallel.

The theorem about parallel planes and lines that is proved in part (a) is theorem 3-1 which states that if two parallel planes are cut by a third plane, then the lines of intersection are parallel.