Točno
30. studenoga 2017. 23:48 (7 godine, 2 mjeseci)
Given three fixed pairwisely distinct points
,
,
lying on one straight line in this order. Let
be a circle passing through
and
whose center does not lie on the line
. The tangents to
at
and
intersect each other at a point
. The segment
meets the circle
at
.
Show that the point of intersection of the angle bisector of the angle
with the line
does not depend on the choice of the circle
.














Show that the point of intersection of the angle bisector of the angle



Upozorenje: Ovaj zadatak još niste riješili!
Kliknite ovdje kako biste prikazali rješenje.
Kliknite ovdje kako biste prikazali rješenje.
Neka je
sjecište simetrale i dužine
. Kao prvi korak primjenimo poučak o simetrali kuta na trokut
: 
Ostaje samo prepoznati da je
po samoj konstrukciji simedijana u trokutu
te mora vrijediti 
Što ukazuje da su nam svi omjeri fiksne vrijednosti koje ovise samo o međusobom položaju točakla
,
, i
pa tako i
ne ovisi o izboru kružnice




Ostaje samo prepoznati da je



Što ukazuje da su nam svi omjeri fiksne vrijednosti koje ovise samo o međusobom položaju točakla




