Question

AP and BQ are the bisectors of two alternate interior angles formed by the intersection of a transversal t with parallel lines $l$ and $m$. If $\angle PAB=x\angle QBA$. Find $x$.

Answer 1

Given that,

Line L and m are parallel , line t is transversal line which cuts l and m at point A and B.

Line PA and QB bisect angle HAB and angle GAQ

$\angle PAB=x.\angle QBA$ ……(1)

Solution-

$\angle HAB=\angle GBA$ (Alternate interior angle) (2)

$\angle HAP+\angle PAB=\angle HAB$ ………(3)

$\angle HAP=\angle PAB$ (Line PA bisect angle HAB) ……..(4)

Similarly, $\angle QBA+\angle QBG=\angle GBA$ …..(5)

$\angle QBA=\angle QBG$ (QB bisect angle GAQ) ……….(6)

Now, form equation (2),(3) and (5),

$\angle HAB=\angle GBA$

$\angle HAP+\angle PAB=\angle QBA+\angle QBG$

From equation (4) and (5) put the value of angle HAP and angle QBG, we get

$2\angle PAB=2\angle QBA$

$\angle PAB=\angle QBA$ ……… (7)

On comparing equation (1) and (7), we get

$x=1$

Hence, this is the answer.