In an isosceles triangle ABC AB = AC. line PF meets lateral sides AB and AC at points P and F, respectively. the length of the segment AP is 31 dm. angle B of triangle ABC is equal to angle APF. find the length of the segment AP
Since, by condition, the angle ABC = APF, and these are the corresponding angles at the intersection of the secant AB of two straight lines BC and PF, these lines will be parallel.
In an isosceles triangle, the angles at the base are equal, the angle ACB = ABC, then the angle APF = ACB.
Since PF is parallel to BC, then the angle ACB = AFP as the corresponding angles at the intersection of parallel straight lines BC and PF secant AC, then the angle APF = AFP, and the triangle APF is isosceles, and therefore AF = AP = 31 dm.
Answer: The length of the AF segment is 31 dm.
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