Neocijenjeno
20. kolovoza 2017. 12:31 (6 godine, 11 mjeseci)
In a plane the circles
![\mathcal K_1](/media/m/0/5/a/05a00b250f840d61d895f7d9b92f7210.png)
and
![\mathcal K_2](/media/m/5/9/9/599bc1a1fd41389ed94ca196d29d3eef.png)
with centers
![I_1](/media/m/4/1/6/416fe38d243ecd6d66747e4d88b6518d.png)
and
![I_2](/media/m/c/b/0/cb0739259063636b82f7b6b3e9dd3de0.png)
, respectively, intersect in two points
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
and
![B](/media/m/c/e/e/ceebc05be717fa6aab8e71b02fe3e4e3.png)
. Assume that
![\angle I_1AI_2](/media/m/9/c/0/9c0ede57b3e373a6d8ac7da550b2c57c.png)
is obtuse. The tangent to
![\mathcal K_1](/media/m/0/5/a/05a00b250f840d61d895f7d9b92f7210.png)
in
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
intersects
![\mathcal K_2](/media/m/5/9/9/599bc1a1fd41389ed94ca196d29d3eef.png)
again in
![C](/media/m/5/a/b/5ab88f3f735b691e133767fe7ea0483c.png)
and the tangent to
![\mathcal K_2](/media/m/5/9/9/599bc1a1fd41389ed94ca196d29d3eef.png)
in
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
intersects
![\mathcal K_1](/media/m/0/5/a/05a00b250f840d61d895f7d9b92f7210.png)
again in
![D](/media/m/7/0/0/7006c4b57335ab717f8f20960577a9ef.png)
. Let
![\mathcal K_3](/media/m/5/9/5/595d793b62d10dc6877f4191ec240c3d.png)
be the circumcircle of the triangle
![BCD](/media/m/3/e/e/3eefa3e34f78e628cbb5cd3988774661.png)
. Let
![E](/media/m/8/b/0/8b01e755d2253cb9a52f9e451d89ec11.png)
be the midpoint of that arc
![CD](/media/m/8/9/5/895081147290365ccae028796608097d.png)
of
![\mathcal K_3](/media/m/5/9/5/595d793b62d10dc6877f4191ec240c3d.png)
that contains
![B](/media/m/c/e/e/ceebc05be717fa6aab8e71b02fe3e4e3.png)
. The lines
![AC](/media/m/6/4/7/647ef3a5d68f07d59d84afe03a9dc655.png)
and
![AD](/media/m/6/9/6/69672822808d046d0e94ab2fa7f2dc80.png)
intersect
![\mathcal K_3](/media/m/5/9/5/595d793b62d10dc6877f4191ec240c3d.png)
again in
![K](/media/m/e/1/e/e1ed1943d69f4d6a840e99c7bd199930.png)
and
![L](/media/m/f/c/1/fc1ae4eb78da7d1352cbf1f8217ab286.png)
, respectively. Prove that the line
![AE](/media/m/c/e/3/ce31f42a92358c211bccb23e6a92fb55.png)
is perpendicular to
![KL](/media/m/b/1/a/b1ab64b407588444cd365224f9b482b4.png)
.
%V0
In a plane the circles $\mathcal K_1$ and $\mathcal K_2$ with centers $I_1$ and $I_2$, respectively, intersect in two points $A$ and $B$. Assume that $\angle I_1AI_2$ is obtuse. The tangent to $\mathcal K_1$ in $A$ intersects $\mathcal K_2$ again in $C$ and the tangent to $\mathcal K_2$ in $A$ intersects $\mathcal K_1$ again in $D$. Let $\mathcal K_3$ be the circumcircle of the triangle $BCD$. Let $E$ be the midpoint of that arc $CD$ of $\mathcal K_3$ that contains $B$. The lines $AC$ and $AD$ intersect $\mathcal K_3$ again in $K$ and $L$, respectively. Prove that the line $AE$ is perpendicular to $KL$.
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Neka su
![\sphericalangle DAC = \alpha \,\, \sphericalangle ADC = \omega](/media/m/e/e/e/eee76a570e24e910c738fc7df01af09b.png)
. Zbog kuta tangente i tetive
![\Rightarrow \sphericalangle DAB = \sphericalangle ACB](/media/m/f/2/5/f25984b3ba7a7405a1e62caf09bc0a2b.png)
i
![\Rightarrow \sphericalangle ABC = 180\, -\, \sphericalangle BAC + \sphericalangle BCA = 180\, -\, \sphericalangle BAC + \sphericalangle BAD =
180\, -\, \sphericalangle DAC = 180\, -\, \alpha \qquad \textbf{(1)}](/media/m/0/e/2/0e2a84ecfb25e160d70ef8c75ed70b73.png)
Analogno
![\Rightarrow \sphericalangle DBA = 180\, - \, \alpha \qquad \textbf{(2)}](/media/m/c/6/7/c6797e6660c60e96651198a05438f55d.png)
Iz
![\textbf{(1)}](/media/m/3/6/0/360636f6a99eec69bd849eaed0bfe0eb.png)
i
![\Rightarrow \sphericalangle DBC = 2 \alpha = \sphericalangle DEC](/media/m/1/8/a/18ad73c371a17cc06ad38c3c2560965f.png)
a budući da vrijedi i
![|DE| = |EC|](/media/m/6/1/1/611412696d2217dc939405fe63a3a026.png)
imamo da je
![E](/media/m/8/b/0/8b01e755d2253cb9a52f9e451d89ec11.png)
središte opisane kružnice
![\triangle ACD](/media/m/e/0/6/e069ccf2c9b0ea6cedbd0c34246862ee.png)
.
Četverokut
![CDKL](/media/m/6/b/0/6b0edd8ae98e7460bd38899e6ebce18d.png)
je tetivan, pa vrijedi
![\sphericalangle LDC = 180 - \omega \Rightarrow \sphericalangle LKC = \omega](/media/m/4/2/b/42b74253b45124eaae81db51621bafe4.png)
Nadalje
%V0
Neka su $ \sphericalangle DAC = \alpha \,\, \sphericalangle ADC = \omega$. Zbog kuta tangente i tetive $ \Rightarrow \sphericalangle DAB = \sphericalangle ACB$ i $ \sphericalangle BAC = \sphericalangle ADB$ $$ \Rightarrow \sphericalangle ABC = 180\, -\, \sphericalangle BAC + \sphericalangle BCA = 180\, -\, \sphericalangle BAC + \sphericalangle BAD =
180\, -\, \sphericalangle DAC = 180\, -\, \alpha \qquad \textbf{(1)}$$ Analogno $$ \Rightarrow \sphericalangle DBA = 180\, - \, \alpha \qquad \textbf{(2)}$$ Iz $\textbf{(1)}$ i $\textbf{(2)}$ $ \Rightarrow \sphericalangle DBC = 2 \alpha = \sphericalangle DEC$ a budući da vrijedi i $ |DE| = |EC|$ imamo da je $E$ središte opisane kružnice $\triangle ACD$.
Četverokut $CDKL$ je tetivan, pa vrijedi $$ \sphericalangle LDC = 180 - \omega \Rightarrow \sphericalangle LKC = \omega$$ Nadalje $$\sphericalangle AEC = 2 \omega \Rightarrow \sphericalangle EAC = 90 - \omega = \sphericalangle EAK \Rightarrow AE \bot KL \qquad \qquad \textbf{q.e.d}$$