See complete Problem 59

Right and Equilateral Triangles, Midpoints. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 59

Labels:
equilateral,
midpoint,
right triangle,
triangle

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We let BA & BC be the x-axis & y-axis resply. and use standard nomenclature for Yr. ABC by calling EF=p and AD=q. We, therefore, need to prove that 4p^2=q^2. We've A:(c,0),B:(0,0),C;(0,a) and angle ABD=30 deg. Hence, D:(V3a/2,a/2),

ReplyDeleteF:(V3a/4,a/4) and E:(c/2,a/2) since BC=BD=a.

4p^2 =4[(V3a/4-c/2)^2+(a/4-a/2)^2

= (V3a/2-c)^2+(-a/2)^2

=(V3a/2-c)^2+(a/2)^2 -----------(1)

while q^2 = (V3a/2-c)^2 + (a/2)^2 ---(2)

By (1) & (2), it is evident that 4q^2 = q^2 or 4EF^2=AD^2 or EF = AD/2

Ajit: ajitathle@gmail.com

http://geometri-problemleri.blogspot.com/2009/06/problem-27-ve-cozumu.html

ReplyDeletehttp://img402.imageshack.us/img402/4481/problem59.png

ReplyDeleteLet G is the midpoint of CD

Connect BE & EG ( see sketch)

Since E is the midpoint of AC => ∆BEC is isosceles and BE=EC, ∠EBC=∠ECB= α

∠FBE=∠ECG=60- α

Triangle BFE congruence to ∆CGE …. ( Case SAS)

So EF=EG= x

E,G are midpoints of AC & CD => EG= a/2= x