Neocijenjeno
Sept. 25, 2019, 11:34 p.m. (6 years, 11 months)
Neka je $M$ polovište stranice $AC$ u šiljastokutnom trokutu $ABC$ u kojem vrijedi $AB>BC$. Neka je $\Omega $ opisana kružnica trokuta $ ABC$. Tangenta na $ \Omega $ u točkama $A$ i $C$ se sijeku u $P$, a $BP$ i $AC$ se sijeku u $S$. Neka je $AD$ visina u trokutu $ABP$, a $\omega$ opisana kružnica trokuta $CSD$. Pretpostavite da se $ \omega$ i $ \Omega $ sijeku u $K\not= C$. Dokažite: $ \angle CKM=90^\circ $.
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Since 
=
, we have 
, so 
is cyclic, and let it be 
. Let 
be the intersection of 
with 
. That means, that it is enough to prove that 
is the diameter of . Since 
, and we want 
, we want triangles 
and 
to be similar, so we want 
, but that is equivalent of proving that 
. Let 
be the intersection of 
and . By the symmedian theorem we know that 
, so 
is tangent to . But as 
is the diameter of , that means 
. So 
, so .
Since $|AP|$=$|PC|$, we have $\angle PMA=90^{\circ}=\angle PDA$, so $PDMA$ is cyclic, and let it be $\epsilon$. Let $X$ be the intersection of $KM$ with $\omega$. That means, that it is enough to prove that $CX$ is the diameter of $\omega$. Since $\angle MAP=\angle CBA=\angle CXA$, and we want $\angle XAP=\angle PMA$, we want triangles $CAX$ and $MAP$ to be similar, so we want $\angle APM=\angle ACX=\angle AKX=\angle AKM$, but that is equivalent of proving that $K \in \epsilon$. Let $T$ be the intersection of $PK$ and $\omega$. By the symmedian theorem we know that $\angle AKM=\angle TKC=\angle TAC$, so $AT$ is tangent to $\epsilon$. But as $AP$ is the diameter of $\epsilon$, that means $\angle TAP=90^{\circ}$. So $\angle AKM=\angle TAM= 90^{\circ}- \angle MAP=\angle APM$, so $K \in \epsilon$.
Školjka 
polovište stranice 
u šiljastokutnom trokutu 
u kojem vrijedi 
. Neka je 
opisana kružnica trokuta 
i 
se sijeku u 
, a 
i 
. Neka je 
visina u trokutu 
, a 
. Pretpostavite da se 
. Dokažite: 
.