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Prove that a line that divides two sides of a triangle proportionally is
parallel to the third side.
1.
1. Let ABC be a triangle. Segment
DE intersects sides AB and CA,
respectively and divides those
sides proportionally, As such,
ADIAB=AE/AC
2
2.
3. Triangles ADE and ABC are
3.
4.
4.
5. Segments DE and BC must be
to one
another
5. By the Converse of the
Theorem

Prove That A Line That Divides Two Sides Of A Triangle Proportionally Is Parallel To The Third Side 1 1 Let ABC Be A Triangle Segment DE Intersects Sides AB And class=

Sagot :

Answer:

Step-by-step explanation:

If (from the image attached) DE is a line segment that divides the sides AB and AC proportionally in ΔABC then we can prove that line segment DE is parallel to BC using the parallel transversal theorem. Based the parallel transversal theorem, the triangles ABC and ADE are similar therefore angles are congruent and line DE must be parallel to line BC

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