Q. 6

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

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.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
NCERT | Angle Sum Property44 mins
Quiz | Lines and Angles40 mins
Quiz | Lines and Angles43 mins
Lines and Angles45 mins
Quiz | Lines and Angles45 mins
NCERT | Intersecting and Non Intersecting lines)41 mins
NCERT | Discussion on Lines And Angles45 mins
Quiz | Angle Sum Property of Triangle40 mins
NCERT | Lines and Angles Part - 142 mins
Quiz | Imp. Qs. on Lines and Angles30 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses