Question

In triangle ABC, D is a point in AB such that AC = CD = DB. If ∠B = 28°, find the angle ACD.

Answer 1

ΔDBC is an isosceles triangle.

As, Side CD = Side DB

⇒ ∠DBC = ∠DCB    ...[If two sides of a triangle are equal, then angles opposite to them are equal]

And ∠B = ∠DBC = ∠DCB = 28°

As the sum of all the angles of the triangle is 180°

∠DCB + ∠DBC + ∠BCD = 180°

⇒ 28° + 28° + ∠BCD = 180°

⇒ ∠BCD = 180° − 56°

⇒ ∠BCD = 124°

Sum of two non-adjacent interior angles of a triangle is equal to the exterior angle.

⇒ ∠DBC+ ∠DCB = ∠ADC

⇒ 28° + 28° = ∠ADC

⇒ ∠ADC = 56°

Now ΔACD is an isosceles triangle with AC = DC

⇒ ∠ADC = ∠DAC = 56°

Sum of all the angles of a triangle is 180°

⇒ ∠ACD + ∠ADC + ∠DAC = 180°

⇒ ∠ACD + 56° + 56° = 180°

⇒ ∠ACD = 180° − 112°

⇒ ∠ACD = 68°