The figure below shows segments AC and EF which intersect at point B. Segment AF is parallel to segment EC: Two line segments AC and FE intersect at B. Segments AF and EC are parallel to each other. Which of these facts is used to prove that triangle ABF is similar to triangle CBE?
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Solution is as follows,
triangle ABF is similar to angle CBE by AA i.e. Angle-angle test of similarity of two triangles. This can be proved as follows,
angle ABF is equal to angle CBE, as opposite angles are same
Also, angle AFB will be similar to angle CEB as the transverse angles made by two parallel segments are same.
Hence, as the condition of Angle-Angle test is satisfied, triangle ABF and triangle CBE are similar to each other.
triangle ABF is similar to angle CBE by AA i.e. Angle-angle test of similarity of two triangles. This can be proved as follows,
angle ABF is equal to angle CBE, as opposite angles are same
Also, angle AFB will be similar to angle CEB as the transverse angles made by two parallel segments are same.
Hence, as the condition of Angle-Angle test is satisfied, triangle ABF and triangle CBE are similar to each other.
Answered by
3
Answer:
triangle ABF is similar to angle CBE by AA i.e. Angle-angle test of similarity of two triangles. This can be proved as follows,
angle ABF is equal to angle CBE, as opposite angles are same
Also, angle AFB will be similar to angle CEB as the transverse angles made by two parallel segments are same.
Hence, as the condition of Angle-Angle test is satisfied, triangle ABF and triangle CBE are similar to each other.
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