Točno
26. svibnja 2022. 22:22 (2 godine, 1 mjesec)
Let
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
be a trapezoid with parallel sides
![AB > CD](/media/m/c/c/3/cc371178312326b85cdec4f6957f4574.png)
. Points
![K](/media/m/e/1/e/e1ed1943d69f4d6a840e99c7bd199930.png)
and
![L](/media/m/f/c/1/fc1ae4eb78da7d1352cbf1f8217ab286.png)
lie on the line segments
![AB](/media/m/5/2/9/5298bd9e7bc202ac21c423e51da3758e.png)
and
![CD](/media/m/8/9/5/895081147290365ccae028796608097d.png)
, respectively, so that
![\frac {AK}{KB} = \frac {DL}{LC}](/media/m/d/a/d/dadd6a7338962e1602445e0615d5ebdf.png)
. Suppose that there are points
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
and
![Q](/media/m/4/5/c/45ce8d14aa1eb54f755fd8e332280abd.png)
on the line segment
![KL](/media/m/b/1/a/b1ab64b407588444cd365224f9b482b4.png)
satisfying
![\angle{APB} = \angle{BCD}](/media/m/6/0/1/601389a28f353e6340c9a2933b843d06.png)
and
![\angle{CQD} = \angle{ABC}](/media/m/3/c/f/3cf135f493830e60a70abccf242e165d.png)
. Prove that the points
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
,
![Q](/media/m/4/5/c/45ce8d14aa1eb54f755fd8e332280abd.png)
,
![B](/media/m/c/e/e/ceebc05be717fa6aab8e71b02fe3e4e3.png)
and
![C](/media/m/5/a/b/5ab88f3f735b691e133767fe7ea0483c.png)
are concylic.
%V0
Let $ABC$ be a trapezoid with parallel sides $AB > CD$. Points $K$ and $L$ lie on the line segments $AB$ and $CD$, respectively, so that $\frac {AK}{KB} = \frac {DL}{LC}$. Suppose that there are points $P$ and $Q$ on the line segment $KL$ satisfying $\angle{APB} = \angle{BCD}$ and $\angle{CQD} = \angle{ABC}$. Prove that the points $P$, $Q$, $B$ and $C$ are concylic.
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Po obratu Talesa, $AD$, $BC$, i $KL$ sijeku se u jednoj tocki $X$. Zelimo $XC*XB=XQ*XP$
$\angle BPA=\angle BCD$ i $\angle BPA + \angle DQC =180^{\circ}$ se mogu upotrijebiti istovremeno uvodenjem tocke $R$ kao tocka na duzini ${LX}$ takva da je $\angle {CRD}= \angle {BPA}$. Ovo je zgodno jer je onda ${DQCR}$ tetivan i $\angle{CRQ}=\angle{BCQ}$ pa je $XC^2=XR*XQ$. Zbog homoteticnosti je jasno da je ${DR}||{AP}$ i ${CR}||{BP}$, iz druge od te dvije paralelnosti zajedno s prijasnjom jednakosti slijedi $XC*XB=XQ*XP$.
28. svibnja 2022. 19:14 | 11235 | Točno |