# In triangle ABC, the median CD is drawn, which cuts off the isosceles triangle CDB (BD = CD)

**In triangle ABC, the median CD is drawn, which cuts off the isosceles triangle CDB (BD = CD) from it. Find the ACB angle if the ACB angle is 64**

Medina CD ∆ABC divides side AD into two equal parts, AD = DB (by definition of the median of a triangle).

Since AD = BD and BD = CD (by condition), then AD = BD = CD.

Consider ∆ACD.

∆ACD is isosceles because AD = CD. Then ∠A = ∠C = 64 °.

By the theorem on the sum of the angles of a triangle, we obtain: ∠A + ∠C + ∠D = 180 °.

Find ∠D: ∠D = 180 ° – (∠A + ∠C) = 180 ° – (64 ° + 64 °) = 52 °.

∠CDA and ∠CDB are adjacent, then ∠CDB = 180 ° – ∠CDA = 180 ° – 52 ° = 128 °.

Consider ∆CDB.

∆CDB is isosceles (CD = BD), then ∠C = ∠B.

By the theorem on the sum of the angles of a triangle, we obtain: ∠В + ∠C + ∠D = 180 °.

Then ∠C = ∠B = (180 ° – ∠D): 2 = (180 ° – 128 °): 2 = 26 °.

∠ACB = ∠ACD + ∠DCB = 64 ° + 26 ° = 90 °.

Answer: ∠ACB = 90 °.