Math Problem Statement

The measures of the three sides of a triangle are AB = 30 in., AC = 50 in., and BC = 60in.. From a point D on side AB, a line DE is drawn through a point E on side AC such that angle AED is equal to angle ABC. If the perimeter of the triangle ADE is equal to 56 in., find the sum of the lengths of line segments BD and CE.

Solution

To solve this problem, we need to use similar triangles and the properties of a triangle's perimeter. Let's analyze the given triangle ABC\triangle ABC and the smaller triangle ADE\triangle ADE that is formed by the line segment DEDE.

Step-by-Step Solution:

  1. Given information:

    • AB=30AB = 30 in.
    • AC=50AC = 50 in.
    • BC=60BC = 60 in.
    • PADE=AD+DE+AE=56P_{ADE} = AD + DE + AE = 56 in.
    • AED=ABC\angle AED = \angle ABC implies that ADEABC\triangle ADE \sim \triangle ABC (angle-angle similarity).
  2. Determine the similarity ratio: Since ADEABC\triangle ADE \sim \triangle ABC, the sides of ADE\triangle ADE are proportional to the sides of ABC\triangle ABC.

    Let's denote the similarity ratio by kk. Thus, the sides of ADE\triangle ADE can be expressed as:

    • AD=kABAD = k \cdot AB
    • AE=kACAE = k \cdot AC
    • DE=kBCDE = k \cdot BC
  3. Express the perimeter of ADE\triangle ADE in terms of kk: The perimeter of ADE\triangle ADE is given by: AD+DE+AE=kAB+kBC+kACAD + DE + AE = k \cdot AB + k \cdot BC + k \cdot AC

    Substituting the known lengths: k30+k60+k50=56k \cdot 30 + k \cdot 60 + k \cdot 50 = 56

    Simplify this expression: k(30+60+50)=56k (30 + 60 + 50) = 56 k140=56k \cdot 140 = 56

    Solve for kk: k=56140=25k = \frac{56}{140} = \frac{2}{5}

  4. Find ADAD, DEDE, and AEAE: Using k=25k = \frac{2}{5}:

    • AD=kAB=25×30=12AD = k \cdot AB = \frac{2}{5} \times 30 = 12 in.
    • AE=kAC=25×50=20AE = k \cdot AC = \frac{2}{5} \times 50 = 20 in.
    • DE=kBC=25×60=24DE = k \cdot BC = \frac{2}{5} \times 60 = 24 in.
  5. Calculate BDBD and CECE: To find BDBD and CECE, we calculate the remaining parts of sides ABAB and ACAC, respectively.

    • BD=ABAD=3012=18BD = AB - AD = 30 - 12 = 18 in.
    • CE=ACAE=5020=30CE = AC - AE = 50 - 20 = 30 in.
  6. Sum of BDBD and CECE: BD+CE=18+30=48 in.BD + CE = 18 + 30 = 48 \text{ in.}

Conclusion:

The sum of the lengths of line segments BDBD and CECE is 48 inches.


Would you like to ask for details or have any questions?

Here are five related questions for further exploration:

  1. How do you determine the similarity ratio between two similar triangles?
  2. What is the significance of angle similarity in triangles?
  3. Can you find the area ratio of ADE\triangle ADE to ABC\triangle ABC?
  4. What would be the length of BEBE in this problem?
  5. How can you use the similarity ratio to find the lengths of all segments of a triangle?

Tip: Always check if triangles are similar using the angle-angle (AA) criterion, as it can simplify finding unknown lengths.

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Math Problem Analysis

Mathematical Concepts

Geometry
Similar Triangles
Perimeter of Triangles

Formulas

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Theorems

Angle-Angle (AA) Criterion for Similarity

Suitable Grade Level

Grades 9-12