We are given that the angle EDG is equal to α.

Angles that lie on the same side of a straight line, are always supplementary and their sum is equal to 180°. Therefore, the angles EDG and CDE are supplementary and EDG+ CDE=180°.

The angle CDE=180°-α.

Two parallel lines combined with two intersecting lines form 2 similar triangles. We are given that the lines HI and DE are parallel, therefore the triangles ABC and CDE are similar and the angles CDE and ABC are equal and their value is 180°-α (we found that CDE=180°-α).

Note that instead of using similar triangles we can use the statement that two parallel lines combined with two intersecting lines form 2 equal alternate angles. We are given that the lines HI and DE are parallel, therefore the angles CDE and ABC are equal (these angles are alternate) and their value is 180°-α (we found that CDE=180°-α).

We are given that the angle FAH is equal to 40°, therefore the angle BAC is also equal to 40° (the angles are vertical, therefore they are equal).

We are given that AC=BC, therefore the angle BAC=ABC.

We found that the angle BAC=ABC, the angle BAC=40° and the angle ABC=CDE=180°-α. Therefore, the angle ABC=CDE= BAC=FAH=180°-α=40°

We can solve the equation 180°-α=40°, getting α=180°-40°=140°