Q. 6

In Fig. 6.13, BA

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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