## Saturday, March 16, 2013

### Problem 863: Circumscribed trapezoid, Inscribed Circle, Tangent, Metric Relations

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 863.

1. EC=CF,GD=FD
EC//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

2. CF/CD = x/5
FD/CD = x/2

CF/CD + FD/CD = 1
x/5 + x/2 = 1
x = 10/7

3. http://img43.imageshack.us/img43/6426/problem863.png

Note 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

4. prof Radu Ion,Sc Bozioru,BuzauMarch 18, 2013 at 12:18 PM

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