In the trapezoid ABCD AB = CD, AC = AD and ∠ABC = 95 °. Find the CAD angle.

Since AB = CD, then the trapezoid ABCD is isosceles.
Then, by the property of an isosceles trapezoid, ∠ABC = ∠BCD = 95 ° and ∠CDA = ∠DAB.
Recalling that the sum of the angles of a convex n-gon is calculated by the formula (n-2) 180 °, we find that the sum of the angles of the trapezoid is (4-2) 180 ° = 360 °.
Then ∠ABC + ∠BCD + ∠CDA + ∠DAB = 360 °
95 ° + 95 ° + ∠CDA + ∠DAB = 360 °
∠CDA + ∠DAB = 170 °
∠CDA = ∠DAB = 170 ° / 2 = 85 °
Consider the triangle ACD.
Since AC = AD, this triangle is isosceles.
Therefore, by the property of an isosceles triangle, ∠CDA = ∠DCA = 85 °
∠BCA = ∠BCD-∠DCA = 95 ° -85 ° = 10 °
∠CAD = ∠DCA = 10 ° (because they are cross-lying for parallel lines AD and BC).
Answer: 10

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