Question

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

Answer 1

$$\angle ALP + \angle BLP = 180^o$$ [Linear pair]
$$\dfrac {1}{2} \angle ALP + \dfrac {1}{2} \angle BLP = \dfrac {1}{2}180^o$$ [multiplying by $$\dfrac {1}{2}$$]
$$\angle ELP + \angle FLP = 90^o$$ [$$\underset {EL} {\rightarrow}$$ and $$\underset {FL} {\rightarrow}$$ are bisectors of $$\angle ALP anf \angle BLP$$]
$$\angle ELF = 90^o$$ [ELF = SLR = 90 vertically opposite angles]
In $$\Delta SLR, \angle SLR+ \angle LRS+ \angle RSL = 180^o$$ [sum of the angles in triangle is $$180^o$$]
$$90 + \angle LRS \angle RSL = 180$$
$$\angle LRS + \angle RSL = 90^o$$