answersLogoWhite

0


Best Answer

Let ABC be a triangle. Let D and E be the mid points of AB and AC respectively. Then the mid-line theorem states that DEBC and DE = BC/2.


Extend DE beyond E to F such that DE = EF. Since AE = CE, triangles ADE and CEF are equal, making CFAB (or CFBD, which is the same) because, for the transversal AC, the alternating angles DAE and ECF are equal. Also,CF = AD = BD, such that BDFC is a parallelogram. It follows that BC = DF = 2·DE which is what we set out to prove.Conversely, let D be on AB, E on AC, DEBC and DE = BC/2. Prove that AD = DB and AE = CE.This is because the condition DEBC makes triangles ADE and ABC similar, with implied proportion,AB/AD = AC/AE = BC/DE = 2.It thus follows that AB is twice as long as AD so that D is the midpoint of AB; similarly, E is the midpoint of AC.

User Avatar

Wiki User

9y ago

Still curious? Ask our experts.

Chat with our AI personalities

FranFran
I've made my fair share of mistakes, and if I can help you avoid a few, I'd sure like to try.
Chat with Fran
CoachCoach
Success isn't just about winning—it's about vision, patience, and playing the long game.
Chat with Coach
MaxineMaxine
I respect you enough to keep it real.
Chat with Maxine

Add your answer:

Earn +20 pts
Q: Prove the mid-line theorem
Write your answer...
Submit
Still have questions?
magnify glass
imp