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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)



1013163_558778_ans_d3958d45a4304e0694e7cec9606d76ee.png

Mathematics

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