In an isosceles triangle, the angles at the base are equal to 70

In an isosceles triangle, the angles at the base are equal to 70, find the sum of the tangential angles that form the sides of this triangle with the middle line passing through their midpoints.

Since the triangle ABC is isosceles, its angles at the base of the AC are equal, ygo BAC = BCA = 70.

The segment KM, according to the condition, the middle line of the triangle ABC, and therefore KM, is parallel to the base of the AC.

Then the triangle AKMC is an isosceles trapezoid.

The sum of the angles of the trapezoid with its lateral side is 180, then the angle AKM = CMK = (180 – 70) = 110.

Then the sum of the angles AKM + CMK = 110 + 110 = 220.

Answer: The sum of the obtuse angles at the midline is 220.