Q. 6

In Fig. 6.13, BA || ED and BC || EF. Show that ABC + DEF = 180°


Answer :

It is given to us –


BA || ED
BC || EF


To show that - ABC + DEF = 180°


Let us extend DE to intersect BC at G, and EF to intersect BA at H. Then, the figure becomes –


Since, BA || DE


BA || GE


We have two parallel lines BA and GE, and BG is a transversal intersecting BA and GE at points B and G respectively.


ABC = EGC - - - - (i)


Also, BC || EF, and GE is a transversal intersecting BC and EF at points G and E respectively.


EGC = HEG - - - - (ii)


Since GE is a ray standing on the line HF. By linear pair axiom,


HEG + GEF = 180°


EGC + GEF = 180° [From equation (ii)]


ABC + GEF = 180°


ABC + DEF = 180°


Hence, proved.


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