sat-dog

54 The correct answer is . It’s given that in triangle ABC, point D on side AB is connected by a line segment with point E on side AC such that line segment DE is parallel to side BC. It follows that parallel segments DE and BC are intersected by sides AB and AC. If two parallel segments are intersected by a third segment, corresponding angles are congruent. Thus, corresponding angles C and AED are congruent and corresponding angles B and ADE are congruent. Since triangle ADE has two angles that are each congruent to an angle in triangle ABC, triangle ADE is similar to triangle ABC by the angle-angle similarity postulate, where side DE corresponds to side BC, and side AE corresponds to side AC. Since the lengths of corresponding sides in similar triangles are proportional, it DE=AE + = follows that . Since point E lies on side AC, AE CE AC. It’s given that BC AC = + = CE 2AE. Substituting 2AE for CE in the equation AE CE AC yields + = = =162 162 AE 2AE AC, or 3AE AC. It’s given that BC . Substituting for BC 1 and 3AE for AC in the equation DE=AE yields 1D6E2 = 3AE , or 1D6E2 = 3. Multiplying BC AC AE 162 =54 both sides of this equation by yields DE . Thus, the length of line 54 segment DE is . 43 SAT PRACTICE TEST #6 ANSWER EXPLANATIONS