Prove : GF is parallel to EJ.

Question

1. ABCD a square
2. (I), (J) the circles with diameter AB, BC
3. F the point of contact of the tangent to (I) through D distinct of A
4. E the inner point of intersection of DF and (J)
5. G the inner point of intersection of AE and (I).

Prove : GF is parallel to EJ.

Considering the definition of F is quite clear,it's easy to determine where F is(Using coordinate),and then easy to know A,F,J are collinear. Notice the similarity of AFB and BFJ,GEF and AFE.Then you know that AF/FJ=AG/GE=4,and easy to get the answer. To know the coordinate of F,just take notice of DF=DA,tan ADI=tan ADF/2=1/2 Please refer to the picture above.Written according to that.

Am I right?

Answer 1

A simple geometric solution:

Draw DI, DJ and IJ, you get isosceles triangle DIJ. Connect D to B. DB is the height of $\triangle DIJ$.Draw a tangent from D to circle I, this line is symmetric to line DF about DB and passes G which is symmetric to E.It touches circle J at H. Angles HEJ and IGF are symmetric about DB and are equal, that means the rays of them must be parallel, so GF is parallel to EJ.