In figure, AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Determine the value of x.
Let ∠BAD = y, ∠BAC = 3y
∠BDA = ∠BAD = y (As AB = DB)
Now, ∠BAD + ∠BAC + 108° = 180° [Linear Pair]
y + 3y + 108° = 180°
4y = 72°
or y = 18°
Now, In ΔADC
∠ADC + ∠ACD = 108° [Exterior Angle Property]
x + 18° = 180°
x = 90°