Prove that a line that divides two sides of a triangle proportionally is parallel to the third side. be sure to create and name the appropriate geometric figures. (this needs to be filled in below.) let abc be a triangle. segment de intersects sides ab and ca, respectively and divides those sides proportionally. as such, ad/ab = ae/ac. by theangle dae is congruent to angle bac. therefore, triangles ade and abc are corresponding angles of similar triangles are so, angle ade is congruent to angle abc. by the converse of thesegments de and bc must be to one another. you. : )

The answers are parallel transversal theorem, similar, congruent, Parallel transversal theorem, parallel.

Step-by-step explanation:  As shown in the attached figure, in ΔABC, DE is a line segment that divides the sides AB and AC proportionally. We are to prove that DE is parallel to BC.

Therefore, the complete proof is as follows -

Let ABC be a triangle. Segment DE intersects sides AB and CA, respectively and divides those sides proportionally. As such, AD/AB = AE/AC. By the Parallel Transversal Theorem, angle DAE is congruent to angle BAC. Therefore, triangles ADE and ABC are similar. Corresponding angles of similar triangles are congruent. So, angle ADE is congruent to angle ABC. By the Converse of the Parallel Transversal Theorem, segments DE and BC must be parallel to one another.

This completes the proof.

can someone answer this question plz

step-by-step explanation:

6 on both sides and twelve on the top and the bottom

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