If in Fig. 6.11, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m.


It is given to us –


EAB and ABH are a pair of alternate interior angles.


AP and BQ are the bisectors of the angles EAB and ABH respectively.


EAP = PAB, and ABQ = QBH - - - - (i)


Also, AP || BQ


We have to show that l || m.


Now, since t is a transversal intersecting two parallel lines AP and BQ at points A and B,


PAB = ABQ (alternate interior angles) - - - - (ii)


Now,


EAB = EAP + PAB, and ABH = ABQ + QBH


EAB = 2 × PAB, and ABH = 2 × ABQ [From equation (i)]


EAB = 2 × PAB, and ABH = 2 × PAB


EAB = ABH - - - - (iii)


From the figure and the equation (iii), we can say that –


t is a transversal intersecting lines l and m such that the alternate angles, EAB and ABH are equal.


Thus, it is true that l || m.


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