Answer:
Step-by-step explanation:
- ΔABC is an isosceles triangle with all 3 vertices at the circle perimeter → CC' will pass through the center of the circle
- ∠BAC =∠ABC = 70° (isosceles triangle)
- ∠ACB = 180° - ∠ABC - ∠BAC = 40° (sum of triangle vertices = 180°)
- ∠ACC' = 1/2 × ∠ACB = 20° (isosceles triangle)
- ∠AEC = 180° - ∠ABC = 110° (sum of opposite angles of cyclic polygon = 180°)
- ∠ECB = 180° - ∠AEC = 70° (same side interior angles)
- ∠ECA = ∠ECB - ∠ACB = 30°
- ∠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