*Answer:* 65%.

Let the area of the triangle $ABC$ be $S$ and the areas of triangles $AKM$, $BKN$, $MKN$ and $CMN$ be $S_1$, $S_2$, $S_3$ and $S_4$, respectively (Fig. 30). Then:

* $S_2 + S_3 + S_4 = \frac{2}{3}S$ since the l.h.s. is the area of the triangle $MBC$ while triangles $MBC$ and $ABC$ have equal altitudes and the ratio of the lengths of the corresponding bases is $\frac{2}{3}$;
* $S_1 + S_3 + S_4 = \frac{3}{2}S_4$ since the l.h.s. is the area of the triangle $ANC$ while triangles $ANC$ and $MNC$ have equal altitudes and the ratio of the lengths of the corresponding bases is $\frac{3}{2}$;
* $S_2 = \frac{1}{3}S_3$ because triangles $BNK$ and $MNK$ have equal altitudes and the ratio of the lengths of the corresponding bases is $\frac{1}{3}$;
* $S_1 = \frac{1}{4}S$ because the ratio of altitudes of the triangles $AKM$ and $ABC$ is $\frac{3}{4}$ while the ratio of the lengths of the corresponding bases is $\frac{1}{3}$.

Substituting $S_1$ and $S_2$ from the latter two equations to the former two equations, we obtain the system of equations
$$\begin{cases} \frac{1}{3}S_3 + S_3 + S_4 = \frac{2}{3}S, \\ \frac{1}{4}S + S_3 + S_4 = \frac{3}{2}S_4. \end{cases}$$
After adding $\frac{1}{3}S_4$ to both sides of the first equation and bringing the term with $S$ to the l.h.s., we obtain the equivalent system
$$\begin{cases} \frac{4}{3}(S_3 + S_4) - \frac{2}{3}S = \frac{1}{3}S_4, \\ (S_3 + S_4) + \frac{1}{4}S = \frac{3}{2}S_4. \end{cases}$$
Solving with respect to $S_3 + S_4$ and $S_4$ yields $S_3 + S_4 = 0.65S$. Hence the area of the quadrilateral $MKNC$ equals 65% of the area of the triangle $ABC$.

Denote the area of a figure $\Pi$ by $S_{\Pi}$. From the conditions of the problem, we obtain the following:

* $\frac{S_{ABM}}{S_{ABC}} = \frac{|AM|}{|AC|} = \frac{1}{3}$, whence $S_{ABM} = \frac{1}{3}S_{ABC}$ and $S_{CBM} = \frac{2}{3}S_{ABC}$;
* $\frac{S_{ABK}}{S_{ABM}} = \frac{|BK|}{|BM|} = \frac{1}{4}$, whence $S_{ABK} = \frac{1}{4} \cdot \frac{1}{3}S_{ABC} = \frac{1}{12}S_{ABC}$ and, consequently, $S_{AMK} = (\frac{1}{3} - \frac{1}{12})S_{ABC} = \frac{1}{4}S_{ABC}$;
* $\frac{S_{BCK}}{S_{BCM}} = \frac{|BK|}{|BM|} = \frac{1}{4}$ (Fig. 31), whence $S_{BCK} = \frac{1}{4} \cdot \frac{2}{3}S_{ABC} = \frac{1}{6}S_{ABC}$ and $S_{MCK} = (\frac{2}{3} - \frac{1}{6})S_{ABC} = \frac{1}{2}S_{ABC}$.

Since $\frac{S_{ABN}}{S_{ACN}} = \frac{|BN|}{|CN|} = \frac{S_{KBN}}{S_{KCN}}$, we obtain
$$\frac{S_{ABN}}{S_{ACN}} = \frac{S_{ABN} - S_{KBN}}{S_{ACN} - S_{KCN}} = \frac{S_{ABK}}{S_{AMK} + S_{MCK}} = \frac{\frac{1}{12}}{\frac{1}{4} + \frac{1}{2}} = \frac{\frac{1}{12}}{\frac{3}{4}} = \frac{1}{9}.$$

Hence $\frac{S_{ABN}}{S_{ABC}} = \frac{S_{ABN}}{S_{ABN}+S_{ACN}} = \frac{S_{ABN}}{S_{ABN}+9S_{ABN}} = \frac{1}{10}$, implying $S_{ABN} = \frac{1}{10}S_{ABC}$.

Altogether, we obtain
$$\frac{S_{MKNC}}{S_{ABC}} = \frac{S_{ABC} - S_{ABN} - S_{AMK}}{S_{ABC}} = 1 - \frac{1}{10} - \frac{1}{4} = 0.65,$$
meaning that the area of the quadrilateral $MKNC$ equals 65% of the area of the triangle $ABC$.