49. Let $h_{a}=\beta_{b}=m_{c}$, where $h_{a}=A P, \beta_{b}=B L$ and $m_{c}=C F-$ are the altitude, bisector, and median drawn from three different vertices of some triangle $A B C$. What value can the ratio of the greatest side of triangle $A B C$ to its smallest side have?
49. First of all, note that if for triangle \(ABC\) the inequality \(m_{a}b = CA\) holds. This immediately follows from the known formula \(^{2}\) \(m_{a} = \frac{1}{2} \sqrt{2 b^{2} + 2 c^{2} - a^{2}}\), but it can also be proven without calculations. Translate triangle \(ABC\) parallel to vector \(\overline{ED}\) to position \(A' B' C'\) (Fig. 128). Since \(B' D = B E\), by the condition \(B' D = B E > A D\); therefore,

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-159.jpg?height=288&width=433&top_left_y=1025&top_left_x=770)

Fig. 128. thus \(B' P > A P\), where \(P\) is the projection of point \(D\) on \(AB\). Since \(A A' = B B'\), then \(B P > A' P\), and consequently,

\[
\frac{a}{2} = B D > A' D = \frac{b}{2}, \quad \text{or} \quad a > b.
\]

Now suppose triangle \(ABC\) is such that \(h_{a} = \beta_{b} = m_{c}\). Since, by the result of problem 48, \(h_{u} = \beta_{b} \geqslant h_{b}\) and \(h_{a} = m_{c} \geqslant h_{c}\), the height \(h_{a}\)

1) This also follows from the fact that in any other arrangement of points \(P, K\), and \(D\), the problem of constructing triangle \(ABC\) from three segments \(h_{a}, \beta_{a}\), and \(m_{n}\) would be unsolvable, as mentioned later (compare with problem 112 from book [17]).

\(^{2}\) See, for example, Adamar [13], p. 124.
is the largest; therefore, from the equality \(h_{a} = m_{c}\) it follows that \(\angle C < 60^{\circ}\) (see problem 50 below). On the other hand, the median \(m_{c}\) is the smallest in triangle \(ABC\), because \(m_{c} = h_{\alpha} \leqslant m_{\alpha}\) and \(m_{c} = \beta_{b} \leqslant m_{b}\) (see problem 48); therefore, by what has just been proven, \(c\) is the largest side of triangle \(ABC\), and \(C\) is its largest angle. Since the sum of the three angles of a triangle is \(180^{\circ} = 3 \cdot 60^{\circ}\), and since \(\angle C \leqslant 60^{\circ}\), all angles of the triangle are \(60^{\circ}\), triangle \(ABC\) is equilateral, and the ratio of its largest side to its smallest side is 1.