In the given figure, AB || CD. Find the values of x, y and z.

Solution:

$A B \| C D$ and let EF and EG be the transversals.

Now, $A B \| C D$ and EF is the transversal.

Then,

$\angle A E F=\angle E F G \quad$ [Alternate Angles]

$\Rightarrow y=75^{\circ}$

$\Rightarrow y=75$

Also,

$\angle E F C+\angle E F D=180^{\circ} \quad$ [Since CFD is a straight line]

$\Rightarrow x+y=180$

$\Rightarrow x+75=180$

$\Rightarrow x=105$

And,

$\angle E G F+\angle E G D=180^{\circ} \quad[$ Since CFGD is a straight line $]$

$\Rightarrow \angle E G F+125=180$

$\Rightarrow \angle E G F=55^{\circ}$

We know that the sum of angles of a triangle is $180^{\circ}$

$\angle E F G+\angle G E F+\angle E G F=180^{\circ}$

$\Rightarrow y+z+55=180$

$\Rightarrow 75+z+55=180$

$\Rightarrow z=50$

$\therefore x=\mathbf{1 0 5}, y=\mathbf{7 5}$ and $z=\mathbf{5 0}$