Question

In the following figure, BA || ED and BC || EF. Show that ∠ABC + ∠DEF = 180°

Answer 1

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.