Question

Let $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ be two parallel lines and $\overleftrightarrow{PQ}$ be a transversal. Let $\overleftrightarrow{PQ}$ intersect $\overleftrightarrow{AB}$ in L. Suppose the bisector of $\angle ALP$ intersect $\overleftrightarrow{CD}$ in R and the bisector of $\angle PLB$ intersect $\overleftrightarrow{CD}$ in S. Prove that
$\angle LRS+\angle RSL={90}^o$

Answer 1

To proof :

$\angle LRS+\angle RSL=90°$

$\angle QLB=\angle ALP$ (vertically opposite angles)

$\angle ALQ=\angle PLB$ (vertically opposite angles)

$\angle ALQ+\angle QLB=180°$

$\angle ALS+\angle SLQ+\angle QLR+\angle RLB=180°$ $[\angle ALQ=\angle ALS+\angle SLQ\angle QLB=\angle QLR+\angle RLS]$

$2\angle SQL+2\angle QLR=180°$ $[\angle R$$\angle S$ is Angle bisector$]$

$\therefore \angle ALS=\angle SLQ\angle QLR=\angle RLB]$

$2(\angle SLQ+\angle QLR)=180°$

$\angle SLQ+\angle QLR=90°⟹\angle SLR=90°$

In $△SLR$

$\angle SLR+\angle RSL+\angle LRS=180°$

$\angle RSL+\angle LRS=180°-90°=90°$ (proved)