Question

problem 24
given: \\( \triangle \mathrm { jkl } \\) is equilateral;
point \\( p \\) is any point
on \\( \overline { \mathrm { jl } } \\).
prove: \\( \mathrm { kp } < \mathrm { jk } \\), \\( \mathrm { kp } < \mathrm { kl } \\),
\\( \mathrm { kp } < \mathrm { jl } \\)

Explanation:

Step1: Use the property of an equilateral triangle

Since \(\triangle JKL\) is equilateral, \(JK = KL=JL\).

Step2: Analyze the relationship in \(\triangle JKP\)

In \(\triangle JKP\), \(KP\) is a side. By the triangle - side relationship (in a triangle, the length of a side is less than the sum of the lengths of the other two sides. Here, if we consider the degenerate case when \(P\) is on \(JL\) (not coinciding with \(J\) or \(L\)), for non - degenerate triangles formed with \(K\), \(J\), \(P\) or \(K\), \(L\), \(P\), and using the fact that the shortest distance from a point (\(K\)) to a line (\(JL\)) is the perpendicular distance. And for any point \(P\) on \(JL\) (except \(J\) and \(L\)), in \(\triangle JKP\), \(KP\) is a non - hypotenuse side (when considering right - angled triangles formed by the height from \(K\) to \(JL\)). So \(KP \angle JKP\) (when \(P
eq J\)) and using the law of sines \(\frac{KP}{\sin\angle KJP}=\frac{JK}{\sin\angle JPK}\), and \(\sin\angle JPK>\sin\angle KJP\) (since \(\angle JPK + \angle KJP+\angle JKP = 180^{\circ}\) and \(\angle JKP>0\) when \(P
eq J\))).

Step3: Use substitution

Since \(JK = KL\) (equilateral triangle property), substituting \(JK\) with \(KL\) in \(KP < JK\), we get \(KP < KL\). Since \(JK=JL\) (equilateral triangle property), substituting \(JK\) with \(JL\) in \(KP < JK\), we get \(KP < JL\).

Answer:

1. In \(\triangle JKP\), non - hypotenuse side property (or triangle side - angle relationship); 6. \(KP < KL\); 7. Substitution (\(JK = KL\) and \(JK=JL\) from equilateral \(\triangle JKL\))