In ΔABC, the external bisector of ∠ACB meets the line through the point A and parallel to the side BC. Let’s prove that ∠ ADC = 90° - 1/2 ∠ ACB

Given: ΔABC, external angle bisectors of ∠ACB, line parallel to BC passing through point A meets

To prove:

Let CD be the external angle bisector of ∠ACB, i.e., CD is angle bisector of ∠ACE. And let line l be the parallel line to side BC through point A. So line l and CD meets at point D. The figure of the above question is as shown below,

Now in ΔABC,

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 this case,

∠ACE = ∠ABC + ∠BAC………(i)

But it is also given CD is angle bisector of ∠ACE, so

∠ACD = ∠DCE

⇒ ∠ACE = 2∠ACD

Substituting the above value in equation (i), we get

2∠ACD = ∠ABC + ∠BAC……..(ii)

Now in ΔABC,

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

So in this case,

∠BAC + ∠ABC + ∠ACB = 180°

Substituting the value from equation (ii) in above equation, we get

2∠ACD + ∠ACB = 180°

⇒ 2∠ACD = 180° - ∠ACB

Hence proved