In △ABC above, BC is produced to D, |AB| = |AC|, and ∠BAC = 50°. Find ∠ACD.
∴ Triangle ABC is isosceles.
:∠ACB = ∠ABC ∠ACD = 180° - ∠ACB (angle on a straight line) Let ∠ACB = ∠ABC = x .: 2x + 50 = 180° (sum of angles in a triangle) 2x = 130°
x = 65° .: ∠ACB = 65° ==> ∠ACD = 180° - 65° = 115°
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