Triangles that cut off similar triangles, and the SSS and SAS tests

A line parallel to one side of a triangle, meeting the other two sides (or their extensions), cuts off a triangle similar to the original one. Why? In △ABC let D be on AB and E on AC with DE ∥ BC. The angle ∠A is common to both triangles, and ∠ADE = ∠ABC because they are corresponding angles of the parallel lines. Two angles are equal, so △ADE ~ △ABC by the AA postulate. This works in every position: inside the triangle, in an hourglass picture where A lies between the lines, or beyond the base. Then AD/AB = AE/AC = DE/BC. Another useful fact: if two triangles are each similar to a third triangle, they are similar to each other. Always write the letters in matching order and say which angles you used.