In the diagrams below, the points A and B are located anywhere along the rays forming ∠C, whose measure is always 40°. The points A and B are connected by a line segment to form △ABC. The point D is the intersection of the bisectors of the exterior angles at A and B.
What is the measure of ∠ADB?
By the exterior angle theorem:
2x = ∠2 + 40 → ∠2 = 2x – 40
2y = ∠1 + 40 → ∠1 = 2y – 40
∠1 + ∠2 = 2x + 2y – 80
2y = ∠1 + 40 → ∠1 = 2y – 40
∠1 + ∠2 = 2x + 2y – 80
But ∠1 + ∠2 = 180 – 40 = 140
140 = 2x + 2y – 80
2x + 2y = 220 → x + y = 110
z = 180 – (x + y) = 180 – 110 = 70
2x + 2y = 220 → x + y = 110
z = 180 – (x + y) = 180 – 110 = 70
So the measure of ∠ADB = z is 70°, no matter where A and B are placed.