Question

In the given figure, $AB||CD$ prove that $p+q-r={180}^o$

Answer 1

Draw a line $KH$ passing through the point $F$ which is parallel to both $AB$ and $CD$

We know that $KF\parallel CD$ and $FG$ is a transversal
from the figure we know that $\angle KFG$ and $\angle FGD$ are alternate angles

So we get

$\angle KFG=\angle FGD=r^o...\left(i\right)$

We also know that $AE\parallel KF$ and $EF$ is a transversal
from the figure we know that $\angle AEF$ and $\angle KFE$ are alternate angles

So we get

$\angle AEF+\angle KFE={180}^o$

By substituting the values we get

$p^o+\angle KFE={180}^o$

$\angle KFE={180}^o-p^o....\left(ii\right)$

By adding both the equations (i) and (ii) we get

$\angle KFG+\angle KFE={180}^o-p^o+r^o$
from the figures $\angle KFG+\angle KFE$ can be written as $\angle EFG$

$\angle EFG={180}^o-p^o+r^o$

We know that $\angle EFG=q^o$

$q^o={180}^o-p^o+r^o$

It can be written as

$p+q-r={180}^o$

Therefore it is proved that $p+q-r={180}^o$