The medians of the ABC trunks intersect at point O. Through point O, a straight line EF is drawn

| February 15, 2021 | education

The medians of the ABC trunks intersect at point O. Through point O, a straight line EF is drawn, parallel to the AC side. Find ЕF if АС = 15cm.

Let us prove that triangles ABC and EBF are similar.

Angle A is common for triangles.

Since, by condition, EF is parallel to AC, then the angle BEF = ABC as the corresponding angles at the intersection of parallel straight lines AC and EF secant AB.

Then triangles ABC and EBF are similar in two angles.

The medians of similar triangles are referred to as the coefficient of their similarity.

K = ВO / BB1.

The medians, at the point of their intersection, are divided in the ratio of 2/1, then BO = 2 * BB1 / 3.

K = ВO / BB1 = 2/3.

Then EF / AC = 2/3.

EF = 2 * AB / 3 = 2 * 15/3 = 10 cm.

Answer: The length of the segment EF is 10 cm.



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