In the trapezoid ABCD AB = CD, ∠BDA = 10 ° and ∠BDC = 109 °. Find the angle ABD.

bystudin | May 31, 2020 | Education

∠ADC = ∠BDA + ∠BDC = 10 ° + 109 ° = 119 °
Because AB = CD, then the trapezoid ABCD is isosceles.
Then ∠ADC = ∠BAD = 119 ° (by the property of an isosceles trapezoid).
Consider the triangle ABD.
By the theorem on the sum of the angles of a triangle:
180 ° = ∠ABD + ∠BDA + ∠BAD
180 ° = ∠ABD + 10 ° + 119 °
∠ABD = 180 ° -10 ° -119 ° = 51 °
Answer: 51

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