In triangle $ABC$, $AB=14$, $BC=6$, $CA=9$. Point $D$ lies on line $BC$ such that $BD:DC=1:9$. The incircles of triangles $ADC$ and $ADB$ touch side $AD$ at points $E$ and $F$. Find the length of segment $EF$.

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Let's first prove the following statement. If the circle inscribed in triangle $KLM$ touches its side $MK$ at point $P$, then $MP = \frac{1}{2}(MK + ML - KL)$.

Proof. Let $Q$ and $R$ be the points of tangency of the inscribed circle of triangle $KLM$ with sides $ML$ and $KL$ respectively (Fig.1). Then

\[
\begin{gathered}
MQ = MP, KP = KR, LQ = LR \\
KL = KR + LR = KP + LQ = (MK - MP) + (ML - MQ) = \\
= MK + ML - (MP + MQ) = MK + ML - 2MP
\end{gathered}
\]

Therefore, $MP = \frac{1}{2}(MK + ML - KL)$, which is what we needed to prove.

Now let's return to our problem. Let the inscribed circles of triangles $ADC$ and $ABD$ touch the segment $AD$ at points $E$ and $F$ respectively, and let point $D$ lie on segment $BC$ (Fig.2). Then, by the proven statement,

\[
AE = \frac{1}{2}(AD + AC - CD), AF = \frac{1}{2}(AD + AB - BD)
\]

Thus,

\[
\begin{gathered}
EF = |AF - AE| = \left|\frac{1}{2}(AD + AB - BD) - \frac{1}{2}(AD + AC - CD)\right| = \\
= \frac{1}{2}|AD + AB - BD - AD - AC + CD| = \frac{1}{2}|AB - AC - BD + CD| = \\
= \frac{1}{2}\left|AB - AC - \frac{1}{10}BC + \frac{9}{10}BC\right| = \frac{1}{2}\left|AB - AC + \frac{1}{5}BC\right| = \frac{1}{2}\left|14 - 9 + \frac{1}{5} \cdot 6\right| = \frac{49}{10}.
\end{gathered}
\]

Now let point $D$ lie outside segment $BC$ (Fig.3). Then it lies on the extension of segment $BC$ beyond point $B$. Similarly to the previous case,

\[
EF = |AF - AE| = \frac{1}{2}|AB - AC - BD + CD| =
\]

## Answer

4.9 or 5.5.