In an isosceles trapezoid ABCD, the diagonal AC is perpendicular to its lateral side and forms

In an isosceles trapezoid ABCD, the diagonal AC is perpendicular to its lateral side and forms a CAD angle of 25 degrees with the base AD. Find the BAC angle.

The sum of the angles of the triangle ACD is 180 °:

<CAD + <ACD + <CDA = 180 °;

<CAD = 25 ° (by condition);

<ACD = 90 ° (by condition);

25 ° + 90 ° + <CDA = 180 °;

<CDA = 180 ° – 25 ° – 90 ° = 65 °.

The trapezoid is isosceles, therefore, <BAD = <CDA = 65 °.

<ABC and <BAD – internal one-sided angles when sectioning parallel bases of a trapezoid:

<ABC + <BAD = 180 °;

<ABC + 65 ° = 180 °;

<ABC = 180 ° – 65 ° = 115 °.

Answer: <ABC = 115 °.