Answer:
1. Given
2. Corresponding Angles Theorem
3. AA Similarity Theorem
4. Corresponding sides are proportional
5. Segment Addition Postulate
[tex]\textsf{6.}\;\;\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}[/tex]
7. Subtract 1 from both sides
8. Take the reciprocal of both sides
Step-by-step explanation:
The diagram shows triangle ABC with point D located on side AB and point E located on side AC.
We are given that the line passing through points D and E is parallel to line segment BC.
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[tex]\large\textsf{Statement 1:}\quad \boxed{\overleftrightarrow{DC}\parallel \overline{BC}}[/tex]
[tex]\large\textsf{Reason 1:}\qquad \:\:\boxed{\textsf{Given}}[/tex]
Explanation:
We are given that the line passing through points D and E is parallel to line segment BC.
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[tex]\large\textsf{Statement 2:}\quad \boxed{\angle 1 \cong \angle 3, \angle 2 \cong \angle 4}[/tex]
[tex]\large\textsf{Reason 2:}\qquad\:\: \boxed{\textsf{Corresponding Angles Theorem}}[/tex]
Explanation:
The Corresponding Angles Theorem states that when two parallel lines are intersected by a transversal, each pair of corresponding angles — those in similar positions — are congruent. In this case, DE and BC are parallel, intersected by transversals AB and AC. Therefore, the angles marked as 1 and 3 correspond, as do the angles marked as 2 and 4.
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[tex]\large\textsf{Statement 3:}\quad \boxed{\Delta ADE \sim \Delta ABC}[/tex]
[tex]\large\textsf{Reason 3:}\qquad\:\: \boxed{\textsf{AA Similarity Theorem}}[/tex]
Explanation:
The AA (Angle-Angle) Similarity Theorem states that if two triangles have two corresponding angles that are congruent, then the triangles are similar. As angles 1 and 2 of triangle ADE are congruent to angles 3 and 4 of triangle ABC respectively, triangle ADE is similar to triangle ABC by the AA Similarity Theorem.
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[tex]\large\textsf{Statement 4:}\quad \boxed{\dfrac{AB}{AD}=\dfrac{AC}{AE}}[/tex]
[tex]\large\textsf{Reason 4:}\qquad\:\: \boxed{\textsf{Corresponding sides are proportional}}[/tex]
Explanation:
In similar triangles, corresponding sides — those in the same relative positions — are proportional. For triangles ABC and ADE, side AB corresponds to side AD, and side AC corresponds to side AE. Therefore, we can set up the proportion AB/AD = AC/AE.
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[tex]\large\textsf{Statement 5:}\quad \boxed{\dfrac{AD+DB}{AD} = \dfrac{AE+EC}{AE}}[/tex]
[tex]\large\textsf{Reason 5:}\qquad\:\: \boxed{\textsf{Segment Addition Postulate}}[/tex]
Explanation:
The Segment Addition Postulate states that for any three points M, N and P on a line, the sum of the lengths of segments MN and NP equals the length of segment MP.
In this case, since point D lies on line segment AB, then AD + DB = AB. Similarly, as point E lies on line segment AC, we have AE + EC = AC. Therefore, we can substitute AD + DB for AB and AE + EC for AC in the proportion from the previous statement.
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[tex]\large\textsf{Statement 6:}\quad \boxed{1+\dfrac{DB}{AB} = 1+\dfrac{EC}{AE}}[/tex]
[tex]\large\textsf{Reason 6:}\qquad\:\: \boxed{\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}}[/tex]
Explanation:
Here, fraction decomposition has taken place, where the fraction has been split into a sum of fractions. A fraction whose numerator is a sum can be expressed as the sum of fractions, where each fraction has one of the terms from the original numerator over the same denominator:
[tex]\dfrac{AD+DB}{AD} = \dfrac{AE+EC}{AE}\\\\\\ \dfrac{AD}{AD}+\dfrac{DB}{AD} = \dfrac{AE}{AE}+\dfrac{EC}{AE} \\\\\\1+\dfrac{DB}{AD} = 1+\dfrac{EC}{AE}[/tex]
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[tex]\large\textsf{Statement 7:}\quad \boxed{\dfrac{DB}{AD}=\dfrac{EC}{AE}}[/tex]
[tex]\large\textsf{Reason 7:}\qquad \:\: \boxed{\textsf{Subtract 1 from both sides}}[/tex]
Explanation:
The Subtraction Property of Equality states that subtracting the same quantity from both sides of an equation maintains equality. In this case, 1 has been subtracted from both sides of the equation from the previous statement.
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[tex]\large\textsf{Statement 8:}\quad \boxed{\dfrac{AD}{DB}=\dfrac{AE}{EC}}[/tex]
[tex]\large\textsf{Reason 8:}\qquad\:\: \boxed{\textsf{Take the reciprocal of both sides}}[/tex]
Explanation:
The reciprocal of a fraction is formed by swapping the numerator and denominator of the original fraction. Multiplying the original fraction by its reciprocal always results in 1. So, in this case, we have taken the reciprocals of the fractions from the previous statement.
