The outer angle at the base of an isosceles triangle is 110 degrees. a) find all the corners of the triangle.

The outer angle at the base of an isosceles triangle is 110 degrees. a) find all the corners of the triangle. b) prove that the midpoint of the base of an isosceles triangle is equidistant from the lines containing the lateral sides

The angle ACB is adjacent to the outer angle of the ВСD, the sum of which is 1800, then the angle ACB = (180 – 110) = 70.
Since the triangle ABC is isosceles, the angle BAC = ACB = 70.
The sum of the inner angles of the triangle is 180, then the angle ABC = (180 – 70 – 70) = 40.
Point H is the middle of AC, then AH = CH.
The segments HK and HM are perpendiculars to the lateral sides, then the triangles ACН and CMН are rectangular in which the angle A = C, the hypotenuse AH = CH, then these triangles are equal in hypotenuse and acute angle, which means HK = HM, which was required to prove.
Answer: The angles of the triangle ABC are 40, 70, 70.