Neocijenjeno
7. listopada 2023. 22:42 (9 mjeseci, 1 tjedan)
Let
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
be an acute triangle with
![D, E, F](/media/m/e/c/a/ecac01ab91092791814283e516ec0b5a.png)
the feet of the altitudes lying on
![BC, CA, AB](/media/m/c/f/7/cf7216e218ef8afb7af9284210a1d7d0.png)
respectively. One of the intersection points of the line
![EF](/media/m/f/5/5/f5594d5ec47ea777267cf010e788fedd.png)
and the circumcircle is
![P.](/media/m/9/5/c/95c3cecafab8ed471d070cfcc15c8428.png)
The lines
![BP](/media/m/e/e/f/eefb4fe46ab8d85b7067c29b24aa4cfc.png)
and
![DF](/media/m/3/d/d/3dd8b7899102ac0f0d215a5d87897f88.png)
meet at point
![Q.](/media/m/7/8/b/78b8b2b44099003e844b852726990bc5.png)
Prove that
![AP = AQ.](/media/m/0/2/f/02f01511d4806c76c85b0b455fb2a0b6.png)
Proposed by Christopher Bradley, United Kingdom
%V0
Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$
Proposed by Christopher Bradley, United Kingdom
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Četverokut $APBC$ je tetivan.
Što znaći da je $\angle APQ = \angle C$
Oznaćimo s $H$ ortocentar trokuta.
Tada je
$$\angle C = \angle EHA = \angle DHB$$
Sad odavde imamo da je
$$\angle HBC = \angle HAC = 90 - \angle C$$
Iz tetivnih četverokuta $AEHF$ i $FHDB$ imamo da je
$\angle EFD = 180 - 2\angle C$ i da je $\angle DFB = \angle C$. Iz toga da je $\angle AFP = \angle DFB$ imamo da je četverokut $AFPQ$ tetivan.
A sad iz $\angle C = \angle AFE = \angle PFB = \angle AQP$ slijedi tvrdnja zadatka.