The ABC triangle is isosceles. the angle ABC is equal to 20 degrees through the vertex B of this triangle a straight line EF is drawn parallel to the straight line AC, find 1) angle ABE 2) angle CBF 3) angle BAC and CB
Since the triangle ABC is isosceles, its angles at the base of the AC are equal.
Angle BAC = BCA = (180 – ABC) / 2 = (180 – 20) / 2 = 80.
According to the condition, the straight line EF is parallel to AC, then the angle ABE = BAC = 80, as the angles lying crosswise at the intersection of parallel straight lines EF and AC secant AC.
Similarly, the angle CBF = BCA = 80.
Answer: 1) Angle ABC = 80, 2) angle CBF = 800, 3) angle BAC = BCA = 80.
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