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 °.