In trapezium ABCD, with base AD – large base. Lines AB and CD meet at point E. Angle AED = 70 °, angle CBE = 35 °. Find angle ABC

To begin with, let’s write down the given problem statement.
Given: ABCD – trapezoid; AD – larger base; angle AED = 70 °; angle CBE = 35 °; ie E – the point of intersection of CD and AB.
Find: corner ADC.
Decision:
ABCD – trapezoid, by convention. AD is a larger base, which means CB is a smaller one. By the trapezoid property, AD is parallel to CB
Angle AED = 70 °; angle CEB = angle AED = 70 ° – as vertical angles.
Now, knowing the angle CBE and the angle CEB, we find the angle ВСЕ:
180 ° – (35 ° + 70 °) = 180 ° – 105 ° = 75 °, by the property of angles in a triangle.
Hence, the angle ВСЕ = 75 °.
Angle ADC = angle DCB – as intersecting angles with parallel lines AD and BC and secant CD.
Since angle DCB = angle ВСЕ = angle ADC = 75 °. Then the angle ADC = 75 °.
Answer: 75 °.