I remember learning this in high school, but I can't track down a proof. Let ABC be a triangle and DE a line segment intersecting the triangle such that D is on AB, E is on AC, and DE is parallel to BC. Then(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\frac{\overline{DE}}{\overline{BC}}=\frac{\overline{AD}}{\overline{AB}}=\frac{\overline{AE}}{\overline{AC}}[/tex]

I duckduckgo'd this as much as I could, but no luck. There is a claim for a proof here

http://ceemrr.com/Geometry1/ParallelSimilar/ParallelSimilar_print.html

and I get the proof for the Triangle Midsegment Theorem, but I don't know what they mean when they say that the Proportional Segments Theorem follows from the Triangle Midsegment Theorem by "repeated application." How so?

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# Proportional Segments Theorem

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