Problem E : 17 499

 Sunt noua zile de cand ALINA mi-a spus acel hotarat...ADIO. De atunci nu cred ca am mai vazut-o pe chat. Poza pe care am pus-o in postare voia sa sugereze ca ma voi lasa de fumat (voiam sa fie o ULTIMA tigara ; EA imi spunea ca nu fumeaza...eu de ce n-as putea ?? Hm, inca nu am reusit...)

In the meantime, we console ourselves with another beauty, Geometry (in Algebra I missed a simple problem)

   "E:17499.   Let  $ABCD$  be a square. We construct the equilateral triangle

           $ABE$  with  $E$  and  $C$  on either side of the line  $AB$  and denote by  $G$  the

 intersection of the lines  $AB\;\; and\;\; CE$ , and by  $F$  the intersection of the 

lines $BD\;\; and\;\; CE$.    We also construct the equilateral triangle  

$DGH$  with $H$  and   $C$ on either side of the line  $DG.$  Show that :

 a) the points  $H,\;\;A\;\; and\;\; F$  are collinear .           b)  $HG=HE.$

{Author : } Adrian BUD, Negrești-Oaș"

Solution CiP

< Figure of the statement>

( to be continued)