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Geometry and Trigonometry Difficulty: Hard

In right triangle $A B C$ , angle $C$ is the right angle and $upper B upper C equals 162$ . Point $D$ on side $A B$ is connected by a line segment with point $E$ on side $A C$ such that line segment $D E$ is parallel to side $B C$ and $upper C upper E equals 2 upper A upper E$ . What is the length of line segment $D E$ ?

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Explanation

The correct answer is $54$ . It’s given that in triangle $A B C$ , point $D$ on side $A B$ is connected by a line segment with point $E$ on side $A C$ such that line segment $D E$ is parallel to side $B C$ . It follows that parallel segments $D E$ and $B C$ are intersected by sides $A B$ and $A C$ . If two parallel segments are intersected by a third segment, corresponding angles are congruent. Thus, corresponding angles $C$ and $A E D$ are congruent and corresponding angles $B$ and $A D E$ are congruent. Since triangle $A D E$ has two angles that are each congruent to an angle in triangle $A B C$ , triangle $A D E$ is similar to triangle $A B C$ by the angle-angle similarity postulate, where side $D E$ corresponds to side $B C$ , and side $A E$ corresponds to side $A C$ . Since the lengths of corresponding sides in similar triangles are proportional, it follows that $\frac{D E}{B C} = \frac{A E}{A C}$ . Since point $E$ lies on side $A C$ , $A E + C E = A C$ . It's given that $upper C upper E equals 2 upper A upper E$ . Substituting $2 upper A upper E$ for $C E$ in the equation $A E + C E = A C$ yields $A E + 2 A E = A C$ , or $3 upper A upper E equals upper A upper C$ . It’s given that $upper B upper C equals 162$ . Substituting $162$ for $B C$ and $3 upper A upper E$ for $A C$ in the equation $\frac{D E}{B C} = \frac{A E}{A C}$ yields $\frac{D E}{162} = \frac{A E}{3 A E}$ , or $\frac{D E}{162} = \frac{1}{3}$ . Multiplying both sides of this equation by $162$ yields $upper D upper E equals 54$ . Thus, the length of line segment $D E$ is $54$ .