If line AC bisects angle BCD, what is the measure in d

We can also get to the answer without any additional constructions.

Referring to the figure drawn below.

Let \(\angle{CDA}\) = x

Therefore \(\angle{GCD}\) = 20 + x (Exterior angle is equal to sum of opposite interior angle)

\(\angle{BCG}\) = \(\angle{GCD}\) = 20 + x (Given AC is the angular bisector of \(\angle{BCD}\))

\(\angle{EBC}\) = \(70^o\) (Sum of angles on a straight line is equal to \(180^o\))

\(\angle{FAC}\) = \(160^o\) (Sum of angles on a straight line is equal to \(180^o\))

Figure EBCAF is a 5 sided polygon, who's sum of angles is = \(540^o\) (Sum of angles of a polygon is (n - 2)*180)

Therefore 90 + 90 + 70 + 160 + \(\angle{FCB}\) = \(540^o\)

\(\angle{C}\) = 540 - 410 = \(130^o\)

Now, \(\angle{FCB}\) + \(\angle{BCG}\) = \(180^o\) (Sum of angles on a straight line is equal to \(180^o\))

130 + 20 + x = 180

x = 180 - 150 = 30

Option B

Arun Kumar

A more intuitive approach:

1. Extend line BC to AD. Let's call this intersection point of BC on AD as "E".
2. Angle AEC = 110 ( corresponding angles) => angle ACE = 50 (180-20-110)
3. Angle BCX(other end of angle bisector) = Angle ACE ( vertically opp.) = 50
4. Since CX is bisector of Angle BCD, angle BCD = 100
5. The larger angle around C must be 360 - smaller angle BCD = 360 - 100 =260
6. Since CA bisects the larger BCD angle as well, angle ACD = 260/2 = 130
7. Angle ADC = 180 - angle CAD - angle ACD = 180 - 20 - 130 = 30. [ Option B]