Neocijenjeno
20. kolovoza 2017. 12:31 (6 godine, 8 mjeseci)
In a plane the circles
and
with centers
and
, respectively, intersect in two points
and
. Assume that
is obtuse. The tangent to
in
intersects
again in
and the tangent to
in
intersects
again in
. Let
be the circumcircle of the triangle
. Let
be the midpoint of that arc
of
that contains
. The lines
and
intersect
again in
and
, respectively. Prove that the line
is perpendicular to
.
%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
. Zbog kuta tangente i tetive
i
Analogno
Iz
i
a budući da vrijedi i
imamo da je
središte opisane kružnice
.
Četverokut
je tetivan, pa vrijedi
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 $$\sphericalangle AEC = 2 \omega \Rightarrow \sphericalangle EAC = 90 - \omega = \sphericalangle EAK \Rightarrow AE \bot KL \qquad \qquad \textbf{q.e.d}$$
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