In triangle ABC, points D and E lie on sides AB and AC respectively such that DE is parallel to BC. If AD = x + 2, DB = 3x + 16, AE = x, and EC = 3x + 5, then the value of x is ____.
Answer (2)
In triangle △ A BC , by the given conditions, line segment D E is parallel to BC . According to the Basic Proportionality Theorem (also known as Thales' theorem), if a line is drawn parallel to one side of a triangle and intersects the other two sides, it divides those sides proportionally.
Let's apply this theorem to triangle △ A BC :
Since D E ∥ BC , we can write: D B A D = EC A E .
We substitute the given segment lengths: 3 x + 16 x + 2 = 3 x + 5 x .
Cross-multiply to eliminate the fractions: ( x + 2 ) ( 3 x + 5 ) = x ( 3 x + 16 ) .
Expand both sides: ( x + 2 ) ( 3 x + 5 ) = 3 x 2 + 5 x + 6 x + 10 = 3 x 2 + 11 x + 10 , x ( 3 x + 16 ) = 3 x 2 + 16 x .
Set the expressions equal to each other: 3 x 2 + 11 x + 10 = 3 x 2 + 16 x .
Subtract 3 x 2 from both sides: 11 x + 10 = 16 x .
Solve for x by isolating it on one side: 10 = 16 x − 11 x , 10 = 5 x . x = 2.
Therefore, the value of x is 2 . This solution shows how to use the Basic Proportionality Theorem to solve for unknown segments in a triangle where a line is parallel to one of its sides.
The value of x in the triangle is 2 , derived using the Basic Proportionality Theorem. By applying the theorem, we set up a proportion and solved for x through cross-multiplication and simplification. This demonstrates how parallel lines affect segment ratios in triangles.
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