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Mon 24 Jan, 2005 08:42 pm
I need help trying to solve this Proof:
Given: Triangle ABC, D is the midpoint of BC
Prove: 2(AB[squared]+AC[squared]) = BC[squared]+4AD[squared]
I am still looking for a way of solving this, your help would greatly be appreciated.
A hard way to do it
This is a really hard way, but it should work. Use Heron's formula to show that the area of ABC is equal to the areas of ABD + ACD. You will end up with three pages of math to simplify the terms, but it's straight algebra.
A better way to do it
OK, here is an easier way.
Place your triangle on a Cartesian grid so that point A is on the origin (0,0), point C is along the X axis (Xc,0) and point B is somewhere in space (Xb,Yb). Using this scheme, point D is at ( (Xb+Xc)/2 , Yb/2 ).
You can now compute the lengths of each segment squared using Pythagoras...
AB^2 = Xb^2 + Yb^2
AC^2 = Xc^2
BC^2 = (Xb-Xc)^2 + Yb^2
AD^2 = [ (Xb+Xc)/2 ]^2 + (Yb/2)^2
Plug into your original conjecture and simplify and all works out.
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