Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

 

Let ABC be a triangle and let a line DE be drawn parallel to side BC, intersecting AB at point D and AC at point E, as shown below:

 

We want to show that BD/AD = CE/AE.

 

Since DE is parallel to BC, we have ∠BDE = ∠BCD and ∠CED = ∠ACB (corresponding angles).

 

Now consider the triangles ABD and AEC. By construction, we have ∠ABD = ∠AEC (they are alternate interior angles formed by the parallel lines DE and BC). Also, ∠ADB = ∠AEC (they are vertical angles). Therefore, the two triangles are similar (by the angle-angle criterion).

 

By the similarity of triangles ABD and AEC, we have:

 

BD/AD = EC/AE

 

which is the desired result. Therefore, if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

 


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