![\Gamma](/media/m/4/e/0/4e08987e1d0700578a2eb5c2fc65dc3b.png)
![d](/media/m/f/7/d/f7d3dcc684965febe6006946a72e0cd3.png)
![\Gamma](/media/m/4/e/0/4e08987e1d0700578a2eb5c2fc65dc3b.png)
![d](/media/m/f/7/d/f7d3dcc684965febe6006946a72e0cd3.png)
![AB](/media/m/5/2/9/5298bd9e7bc202ac21c423e51da3758e.png)
![\Gamma](/media/m/4/e/0/4e08987e1d0700578a2eb5c2fc65dc3b.png)
![AB](/media/m/5/2/9/5298bd9e7bc202ac21c423e51da3758e.png)
![d](/media/m/f/7/d/f7d3dcc684965febe6006946a72e0cd3.png)
![B](/media/m/c/e/e/ceebc05be717fa6aab8e71b02fe3e4e3.png)
![d](/media/m/f/7/d/f7d3dcc684965febe6006946a72e0cd3.png)
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
![C](/media/m/5/a/b/5ab88f3f735b691e133767fe7ea0483c.png)
![\Gamma](/media/m/4/e/0/4e08987e1d0700578a2eb5c2fc65dc3b.png)
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
![B](/media/m/c/e/e/ceebc05be717fa6aab8e71b02fe3e4e3.png)
![D](/media/m/7/0/0/7006c4b57335ab717f8f20960577a9ef.png)
![AC](/media/m/6/4/7/647ef3a5d68f07d59d84afe03a9dc655.png)
![d](/media/m/f/7/d/f7d3dcc684965febe6006946a72e0cd3.png)
![D](/media/m/7/0/0/7006c4b57335ab717f8f20960577a9ef.png)
![\Gamma](/media/m/4/e/0/4e08987e1d0700578a2eb5c2fc65dc3b.png)
![\Gamma](/media/m/4/e/0/4e08987e1d0700578a2eb5c2fc65dc3b.png)
![E](/media/m/8/b/0/8b01e755d2253cb9a52f9e451d89ec11.png)
![B](/media/m/c/e/e/ceebc05be717fa6aab8e71b02fe3e4e3.png)
![E](/media/m/8/b/0/8b01e755d2253cb9a52f9e451d89ec11.png)
![AC](/media/m/6/4/7/647ef3a5d68f07d59d84afe03a9dc655.png)
![F](/media/m/3/e/8/3e8bad5df716d332365fca76f53c1743.png)
![BE](/media/m/e/e/2/ee25cd134664bc0c8d7fdbba81e54f90.png)
![d](/media/m/f/7/d/f7d3dcc684965febe6006946a72e0cd3.png)
![AF](/media/m/a/e/4/ae455e708e936870cb86e6a074a2c5a0.png)
![\Gamma](/media/m/4/e/0/4e08987e1d0700578a2eb5c2fc65dc3b.png)
![G](/media/m/f/e/b/feb7f8fc95cee3c3a479382202e06a86.png)
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
Prove that the reflection of the point
![G](/media/m/f/e/b/feb7f8fc95cee3c3a479382202e06a86.png)
![AB](/media/m/5/2/9/5298bd9e7bc202ac21c423e51da3758e.png)
![CF](/media/m/6/7/0/670c216bc8a05762a60542376587c5fc.png)
Kliknite ovdje kako biste prikazali rješenje.
U rješenju, svi polovi i polare referiraju se na . Neka je
preslika
preko
,
presjek
i
.
je ortocentar u trokutu
, stoga je
kolinearno. (
i
su oba okomiti na
).
:
je tangenta na
.
dokaz: Neka je presjek
i
, te
presjek
i
. Poznato je da
, pa projiciranjem dobivamo:
, odakle slijedi da je
na polari od
, neka je ta polara
. Nadalje, znamo da je
okomit na
, pa je jedina opcija za polaru
.
Sad, La Hire daje da je na polari od
, no kako je trivijalno
na polari od
imamo da je
polara od
, pa je
na toj polari i po definiciji polare slijedi tvrdnja.
Promotrimo četverokut , on je tetivan zbog pravih kuteva, a kako je
slijedi da je
centar opisane mu kružnice, odnosno
je polovište
.
Sad slijedi ključna obzervacija: je
-Humpty točka u
Sad ide završetak zadatka: iz svojstava Humpty točke imamo:
i gotovi smo