In the given figure, AD bisects ∠BAC in the ratio 1: 3 and AD = DB. Determine the value of x.
Let the ratio be y
⸫ ∠DAB = y
⸫ ∠DAC = 3y
⸫ y + 3y + 108° = 180° [Angle on a straight line]
⇒ 4y = 72°
⸫ y = 18°
⸫ ∠DAC = 3y = 54°
∠ABD = 18° [⸪ AD = DB, ΔABD is an isosceles triangle]
In ΔABC,
⇒ x + ∠A + ∠B = 180° [Sum of all angles of a triangle = 180°]
⇒ x = 180° - 72° - 18°
⸫ x = 90°