Segment AK – bisector of triangle CAE. A straight line is drawn through point K, parallel to the side CA and intersecting

Segment AK – bisector of triangle CAE. A straight line is drawn through point K, parallel to the side CA and intersecting the side AE at point N. Find the angles of the triangle AKN if the angle CAE = 92 degrees.

By condition, we are given the parallelism of the CA side and the straight line KN. That is, we can say that these are two straight lines intersected by the secant AE. From this we start. We are given the bisector of the CAE angle. If the entire angle CAE = 92 degrees, then one knot of the angles of our triangle ANK – the angle KAN will be equal to 92: 2. We get one of the corners of the triangle. Angle KAN = 46 degrees. We have already said that lines CA and KN are parallel at secant AE. Then our angle of 92 degrees will be an internal one-sided angle with an angle of ANK. And by the property we know that if the straight lines are parallel, then the sum of the internal one-sided = 180 degrees. Thus, we can find the angle ANK. 180-92 = 88 degrees. Well, it remains to find one corner of the AKN. Knowing that the sum of the angles of a triangle = 180, then: 180- (46 + 88) = 46 degrees. Well, we solved the problem. For prophylaxis, I can say that the triangle from which we were looking for the corners turned out to be isosceles (it has equal angles at the base).