All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
Medially Similar (Posted on 2005-08-19) Difficulty: 2 of 5
Let ABC be a triangle, with M the midpoint of BC, and DEF a triangle, with N the midpoint of EF. Suppose that angle BAM equals angle EDN and angle CAM equals angle FDN. Show that triangles ABC and DEF are similar.

See The Solution Submitted by McWorter    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 3
Clearly, <BAC = <BAM + <CAM = <EDN + <FDN = <EDF.
Using the sine rule,
  |BA| sin(<BAM) = |BM| sin(<BMA)
  |AC| sin(<CAM) = |CM| sin(<AMC)
Since |BM| = |CM| and <BMA + <AMC = 180,
  |BA| sin(<BAM) = |AC| sin(<CAM) 
Using the sine rule,
  |ED| sin(<EDN) = |EN| sin(<END)
  |DF| sin(<FDN) = |FN| sin(<DNF)
Since |EN| = |FN| and <END + <DNF = 180,
  |ED| sin(<EDN) = |DF| sin(<FDN) 
  |ED| sin(<BAM) = |DF| sin(<CAM)
  |BA|     |ED|
 ------ = ------
  |AC|     |DF|
Hence, triangles ABC and DEF are similar.

  Posted by Bractals on 2005-08-19 15:11:03
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information