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Explanation:
To prove that a line that divides two sides of a triangle proportionally is parallel to the third side, we can use the concept of similar triangles. Let's create and name the appropriate geometric figures to aid our explanation.
Consider a triangle ABC, where AB, BC, and AC are the sides of the triangle. Let's draw a line segment DE that intersects AB and BC, dividing them proportionally.
To represent this visually, we can draw triangle ABC and label the vertices as A, B, and C. Then, draw a line segment DE that intersects AB at point D and BC at point E.
Now, let's prove that DE is parallel to AC.
To begin the proof, we need to establish that triangles ADE and CDE are similar.
Using the concept of proportional sides, we can say that:
AD/DB = CE/EB
This is because DE divides AB and BC proportionally.
Next, we need to show that the corresponding angles of triangles ADE and CDE are congruent.
Angle ADE is congruent to angle CDE because they are vertical angles (opposite angles formed by the intersection of two lines).
Finally, we can conclude that triangles ADE and CDE are similar by the Angle-Angle (AA) similarity criterion.
Since triangles ADE and CDE are similar, their corresponding sides are proportional.
Now, let's focus on the sides AD and CE.
Since AD/DB = CE/EB, we can rewrite this as AD/CE = DB/EB.
By the transitive property of proportions, we can say that AD/CE = AB/BC.
This implies that the ratio of the lengths of the corresponding sides of triangles ADE and CDE is equal to the ratio of the lengths of the corresponding sides of triangle ABC.
Therefore, by the Converse of the Corresponding Angles Postulate, we can conclude that DE is parallel to AC.
We have proven that a line that divides two sides of a triangle proportionally is parallel to the third side by establishing the similarity of triangles ADE and CDE and using the corresponding sides proportionality.
For further reference concerning Concept Of Similar Triangles; brainly.com/question/30850803
Step-by-step explanation:
Assume that we have triangle ABC and a line DE parallel to the side BC, where D is on AB and E is on AC. Now extend DE to meet BC at F. Using the Intercept theorem, we can show that DE is parallel to BC.
Triangles ADE and ABC share a common side AD. According to the Intercept Theorem since DE is parallel to BC, the segments DF and FC divide the side AB proportionally just like DE divides BC proportionally. Therefore DE is parallel to BC.
In triangle ABC a line DE is parallel to side BC and dividing sides AB and AC proportionally implies DE is parallel to the third side BC.
