Given:
DH = HE
m(ar BG) = (9x - 20)°
m(ar GC) = (5x + 28)°
To find:
The m(ar AB)
Solution:
Radius bisects the chord and its arc.
m(ar BG) = m(ar GC)
(9x - 20)° = (5x + 28)°
9x° - 20° = 5x° + 28°
Add 20° on both sides.
9x° - 20° + 20° = 5x° + 28° + 20°
9x° = 5x° + 48°
Subtract 5x° from both sides.
9x° - 5x° = 5x° + 48° - 5x°
4x° = 48°
Divide by 4 on both sides, we get
x° = 12°
Substitute x = 12 in m(ar BG) and m(ar GC).
m(ar BC) = m(ar BG) + m(ar GC)
= (9x - 20)° + (5x + 28)°
= (9(12) - 20)° + (5(12) + 28)°
= 88° + 88°
m(ar BC) = 176°
Chord AB congruent to chord BC.
m(ar AB) = m(ar BC)
m(ar AB) = 176°
The measure of arc AB is 176°.