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 diagram below for more details.
Let u= ∠(BCE) and t= tan(u)We have ET=x.tan(u)=3 ..... (1)DG= x/cos(90-2.u)-x= x/sin(2u)-xAnd ET/DG=3= tan(u)/(1/sin(2u)-1)Replace sin(2u)=2t/(t^2+1) in above We get 2.t^2/(t-1)^2=3 so t= √ (6)/( √ (6)+2)Replace value of t in (1) we get x= 3+√ (6)
Let the intersection of EG & AD be called F and let FT=FD=y. From similar triangles FEA & FDG we can say that (x-3)/1=(x-y)/y and right triangle FEA gives us, (3+y)²=(x-3)²+(x-y)². Solving these two equations, we obtain the only admissible value of x as (3 +√6)
Purely Geometry solutionLet TG = pThen from Pythagoras p^2 + x^2 = (x+1)^2So p^2 = 2x+1. (1)Now write the area of Tr. CEG in 2 ways1/2 x(p+3) = 1/2 x (x+1)So p = x - 2. and from (1)(X-2)^2 = 2x + 1 so x^2 -6x + 3 = 0Solving the quadratic x = 3 + or - sqrt 6Sumith PeirisMoratuwa Sri Lanka
Alternatively since 2 altitudes of the Tr. ECG are equal the corresponding sides can easily be proved to be equalHence x+1 = p+3 and so on
Note that x=3+√ (6) is only acceptable solution of this problem.Peter Tran
EC bisects angle BET (BEC)So angles GCE = CEB = CEGThus GE = GC = 1 + x, GT = GE - 3 = x - 2Let GC be extended to meet circle at D'So GT^2 = GD. GD' = 1.(1 + 2x) = 1 + 2x.Follows (x- 2)^2 = 2x + 1, x^2 - 6x + 3 = 0 etc
I use the point F like the user Ajit did. My solution here:https://www.facebook.com/photo.php?fbid=10206223938546060&set=a.10205987640598759.1073741831.1492805539&type=3&theater
Excellent observation of Sumith Peiris, GC=GE; now we can apply the power of G w.r.t. circle, GT^2=GD(GD+2CD), so (x-2)^2=2x+1, a.s.o.