Question

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

Solution

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$$]$$$$\therefore \angle ALS=\angle SLQ\angle QLR=\angle RLB]$$$$2(\angle SLQ+\angle QLR)=180°$$$$\angle SLQ+\angle QLR=90°\Longrightarrow \angle SLR=90°$$In $$\triangle SLR$$$$\angle SLR+\angle RSL+\angle LRS=180°$$$$\angle RSL+\angle LRS=180°-90°=90°$$   (proved)Mathematics

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