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Answer:

Step-by-step explanation:

  1. ΔABC is an isosceles triangle with all 3 vertices at the circle perimeter → CC' will pass through the center of the circle
  2. ∠BAC =∠ABC = 70° (isosceles triangle)
  3. ∠ACB = 180° - ∠ABC - ∠BAC = 40° (sum of triangle vertices = 180°)
  4. ∠ACC' = 1/2 × ∠ACB = 20° (isosceles triangle)
  5. ∠AEC = 180° - ∠ABC = 110° (sum of opposite angles of cyclic polygon = 180°)
  6. ∠ECB = 180° - ∠AEC = 70° (same side interior angles)
  7. ∠ECA = ∠ECB - ∠ACB = 30°
  8. ∠DCC' = ∠DCE + ∠ECA + ∠ACC' = 90°

Since CC' pass through the center of the circle (coincide with the diameter) and ∠DCC' = 90°. Therefore DC is perpendiculer with CC' and tangent with the circle at C

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