Question

in the given figure AB parallel DE Prove that angle ABC + angle BCD= 180° + angle CDE

Answer 1

here is your answer user

Answer 2

Answer:

We have to prove that the ∠ABC+∠BCD=180°+∠CDE by using the properties of intersection of lines.

• Sum of co-interior angles are 180°
• Alternate angles are the angles formed by the two parallel lines crossed by a transversal are  equal.

Step-by-step explanation:

CF||BA and CB is the transversal

∠ABC+∠BCF=180°  ..............................(i) (co-interior ∠s)

Then ,FC||DE and DC is the transversal.

∠FCD=∠CDE...(ii) (alternate interior ∠s ).

Adding the corresponding sides of (i) and (ii),

we get ∠ABC+(∠BCF+∠FCD)=180°+∠CDE

                                               ⇒∠ABC+∠BCD

                                                =180°+∠CDE.