See complete Problem 228 at:
gogeometry.com/problem/p228_triangle_midpoint_perpendicular.htm
Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Tuesday, January 20, 2009
Elearn Geometry Problem 228: Triangle, Midpoint, Exterior line
Subscribe to:
Post Comments (Atom)
We have:
ReplyDeletein the trapezoid AA'B'B : d=(a+b)/2
in the trapezoid BB'C'C : e=(b+c)/2
in the trapezoid AA'C'C : f=(a+c)/2
From this relations we have :
d+e+f=(2a+2b+2c)/2=a+b+c
See the drawing
ReplyDelete- DD’ intersects AC in D’’, BB’ in B’’, EE’ in E’’
- Define d1=DD’’, d2=D’’D’ with d=d1+d2
- Define b1=BB’’, b2=B’’B’ with b=b1+b2
- Define e1=EE’’, e2=E’’E’ with e=e1+e2
- F in the middle of AC =>FF’=f=(a+c)/2
- DAD’’ is similar to BAB’’ =>BA/DA=b1/d1
- BA/DA=2=> d1=b1/2
- In the same way, e1=b1/2=d1
- DAD’’ is similar to BAB’’ =>BA/DA=B’’A/D’’A=2 => D’’ A=D’’B’’
- In the same way, E’’B’’=E’’C
- =>d2=(a+b2)/2 and e2=(c+b2)/2
- d+f+e=d1+d2+f+e1+e2=b1/2+(a+b2)/2+(a+c)/2+(b1/2)+(c+b2)/2
- =(b1+a+b2+a+c+b1+c+b2)/2=(2b+2a+2c)/2
- =>d+f+e=a+b+c