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Rule

Converse Triangle Proportionality Theorem

If a segment is drawn between two sides of a triangle such that it divides the sides proportionally, the segment is parallel to the third side in the triangle.

Based on the diagram, the following relation holds true.

If $\frac{A D}{D C} = \frac{B E}{E C},$ then $\overline{D E} ∥ \overline{A B} .$

Proof

The given proportion can be rearranged to get the proportionality of two sides of

△ A B C

and

△ D E C .

\frac{A D}{D C} = \frac{B E}{E C}

▼

Rewrite

AddEqn

$LHS + 1 = RHS + 1$

\frac{A D}{D C} + 1 = \frac{B E}{E C} + 1

OneToFrac

Rewrite $1$ as $\frac{a}{a}$

\frac{A D}{D C} + \frac{D C}{D C} = \frac{B E}{E C} + \frac{E C}{E C}

AddFrac

Add fractions

\frac{A D + D C}{D C} = \frac{B E + E C}{E C}

Segment Addition Postulate

\frac{A C}{D C} = \frac{B C}{E C}

This means that

△ A B C

is a dilation of

△ D E C

from point

C

with scale factor

r = \frac{A C}{D C} = \frac{B C}{E C} .

A dilation moves a segment to a parallel segment, so the proof is complete.

If $\frac{A D}{D C} = \frac{B E}{E C},$ then $\overline{D E} ∥ \overline{A B} .$

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Proof