Question

The side BC of ΔABC is produced on both sides.
Prove the sum of the exterior angles so formed is greater than angle A by two right angles.​

Answer 1

Answer:

If we extend BC on both ends, then the angles would be,

∠ACD=180−C

∠ABF=180−B

Sum of the 2 exterior angles=∠ACD+∠ABF

=(180−C)+(180−B)

=360−(B+C)

=360−(180−A)=180+A

∠A is always positive.

∴ sum of the angles is greater than 180°.

Answer 2

Given :-

• The side BC of ΔABC is produced on both sides to point D and E respectively.

To Prove :-

• The sum of the two exterior angles formed is greater than ∠ A by two right angles i.e. 180°

Proof :-

According to linear pair :

• ∠ABE = 180° - ∠ABD { eq. I }
• ∠ACD = 180° - ∠ACE { eq. ii }

Now,

=> ∠A + ∠B + ∠C = 180° {Angle Sum Property of a Triangle}

Now we will put the values of ∠B and ∠C from eq. (I) and eq. (ii) we get ,

=> (180° - ∠ABD) + (180° - ∠ACE) + ∠A = 180°

=> 360° + ∠ABD + ∠ACE = 180° - ∠A

=> ∠ABD + ∠ACE = ∠A + 180°

We know that angles of a Triangle are always positive.

Hence Proved !