If we produce the side BC on both sides, the two exterior angles are formed. Let’s prove that the sum of the measurement of these two exterior angles is more than 2 right angles.

Given: ΔABC, exterior angles ∠ABD and ∠ACE

To prove: the sum of the measurement of these two exterior angles is more than 2 right angles, i.e., ∠ABD + ∠ACE > 2(90°)

The figure for the given question is as shown below,

We know in a triangle the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

So in ΔABC ∠ABD and ∠ACE are exterior angles, so

∠ABD = ∠BAC + ∠ACB……..(i)

And,

∠ACE = ∠ABC + ∠BAC……..(ii)

Adding equation (i) an equation (ii), we get

∠ABD + ∠ACE = ∠BAC + ∠ACB + ∠ABC + ∠BAC …….(iii)

We also know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ACB + ∠ABC = 180°

Substituting the above equation in equation (iii), we get

∠ABD + ∠ACE = (∠BAC + ∠ACB + ∠ABC) + ∠BAC

⇒ ∠ABD + ∠ACE = (180°) + ∠BAC

⇒ ∠ABD + ∠ACE > 180°

⇒ ∠ABD + ∠ACE > 2(90°)

Hence the sum of the measurement of these two exterior angles is more than 2 right angles.

Hence proved