Problem 93. Which distance is greater: from the vertex of the triangle to the incenter or from the incenter to the base of the bisector.
Solution. The property $\frac{A I}{I L_{1}}=\frac{b+c}{a}$ and the triangle inequality immediately give the answer: $A I > I L_{1}$.

Problem 94 (Kiev City Olympiads, 1987). The incircle of triangle $A B C$ touches side $B C$ at point $K$. Prove that segment $A K$ is shorter than the diameter of the circle.

Proof. Draw the altitude

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-075.jpg?height=497&width=645&top_left_y=845&top_left_x=337)

$A H_{1}$ and segment $I T$ parallel to $B C$ (Fig. 99). $T H_{1}=r ; A T=h_{a}-r$. It is obvious that:

$$
\frac{A I}{I L_{1}}=\frac{A T}{T H_{1}}=\frac{h_{a}-r}{r}=\frac{h_{a}}{r}-1
$$

But by the property of the incenter

$$
\frac{A I}{I L_{1}}=\frac{b+c}{a} \text {. }
$$

Thus, $\frac{b+c}{a}=\frac{h_{a}}{r}-1$. But $b+c>a$, i.e., $\frac{b+c}{a}>1$. Therefore, $\frac{h_{a}}{r}-1>1$, or $h_{a}>2 r$. Since $A K>h_{a}$, then $A K>2 r^{1}$.

Problem 95 (MFTI Olympiad).

The bisector $A_{4}$ intersects the incircle of triangle $A B C$ at points $E$ and $T$ (Fig. 100). Which segment is longer: $A E$ or $T L_{1}$?

Solution. $A E=A I-r, T L_{1}=I L_{1}-r_{1}$ $\frac{A I}{I L_{1}}=\frac{b+c}{a}$ and $A I>I L_{1}$. Then: $A I-r>I L_{1}-r$, i.e., $A E>T L_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-075.jpg?height=508&width=674&top_left_y=1645&top_left_x=1062)[^2]

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