Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view the complete problem 1043.
C, I, E are collinearConnect I with E and A with B AE = AI = 3√2in the triangle AIC we have IC = 3√3 - 3triangle EBC similar to triangle AIC, EB/AI = EC/ACx = 3(1 + √3)/2x= 4.098Erina, NJ
Radius of incircle=r. Drop I onto ED to obtain point P. AB=x+r=PI, so triangle PIE is congruent to triangle BAC. BC=PE=x-r. DC=2x-r=6+r. Therefore1) x-r=3.Substitute this into (x-r)^2+(x+r)^2=36 in triangle PIE to get 2) x+r=3*sqrt(3).Adding 1) and 2) yields 2x=3(1+sqrt(3))==>x=3(1+sqrt(3))/2