Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 863.
Saturday, March 16, 2013
Problem 863: Circumscribed trapezoid, Inscribed Circle, Tangent, Metric Relations
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EC=CF,GD=FD
ReplyDeleteEC//GD ,so EK/KD=EC/GD , or EK/KD=CF/FD therefore KE//EC
x/EC=DF/DC so, x/2=5/7 ,x=10/7
CF/CD = x/5
ReplyDeleteFD/CD = x/2
CF/CD + FD/CD = 1
x/5 + x/2 = 1
x = 10/7
http://img43.imageshack.us/img43/6426/problem863.png
ReplyDeleteNote that Triangle ECK similar to ∆DGK ( case AA)
We have CF/FD=CE/DG=CK/KG
So KG//DG and ∆CKF similar to ∆CGD ( case SAS)
From ∆CKF similar to ∆CGD we have
DG/FK=5/x=CG/CK=(CK+KG)/CK= 1+KG/CK= 7/2
So x=10/7
Fie ABCD un trapez cu bazele AB=a,CD=b in care notam cu O intersectia diagonalelor.Daca notam cu M si N intersectia paralelei prin O la baze cu laturile neparalele avem ca MN este media armonica a bazelor ,MN=2ab/a+b si ON=OM=MN/2 in general,In cazul particular al problemei 863 In trapezulECDG a=2,b=5 si KF=2.2.5/2(2+5)=10/7
ReplyDeleteTriangle CKE ~ Triangle GKD
ReplyDeleteCK/KG=CE/GD=2/5
CK/CG=CK/(CK+KG)=2/7
Triangle CKF ~ Triangle CGD
KF/GD=CK/CG=2/7
x/5=2/7
x=10/7