MHB Equilateral Triangle Intersecting Lines Theorem

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In an equilateral triangle ABC with an interior point K, the lines AK, BK, and CK intersect the sides BC, CA, and AB at points A', B', and C', respectively. The theorem states that the product of the segments A'B', B'C', and C'A' is greater than or equal to the product of the segments A'B, B'C, and C'A. This inequality highlights a geometric relationship involving the segments created by the intersection of lines from an interior point to the triangle's vertices. The proof involves analyzing the properties of equilateral triangles and the relationships between the segments formed. The discussion emphasizes the significance of this theorem in understanding geometric properties and relationships within equilateral triangles.
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Let $ABC$ be an equilateral triangle, and let $K$ be a point in its interior. Let the line $AK,\,BK,\,CK$ meet the sides of $BC,\,CA,\,AB$ in the points $A',\,B',\,C'$ respectively. Prove that

$A'B'\cdot B'C'\cdot C'A' \ge A'B\cdot B'C\cdot C'A$.
 
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Solution of other:

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Let $A'B=a,\,A'C=a',\,B'C=b,\,B'A=b',\,C'A=c$ and $C'B=c'$. Then by Ceva's Theorem, we have $abc=a'b'c'$---(*).

Since $\angle B'AC'=60^{\circ}$, we have

$\begin{align*}B'C'^2&=(b')^2+c^2-2b'c\cos 60^{\circ}\\&=(b')^2+c^2-b'c\ge b'c\end{align*}$

Similarly we have

$C'A'\ge c'a$ and $A'B'^2\ge a'b$

Multiplying these three inequalities, we get

$B'C'^2\cdot C'A'^2\cdot A'B'^2\ge b'c\cdot c'a \cdot a'b$---(**)

From (*) and (**) we have $B'C'^2\cdot C'A'^2\cdot A'B'^2 \ge a^2b^2c^2$

Thus we have $B'C'\cdot C'A'\cdot A'B' \ge abc$

That is, $A'B'\cdot B'C'\cdot C'A' \ge A'B\cdot B'C\cdot C'A$.
 

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