Problem 1719

To find the measure of angle $\angle CAB$ in triangle $\triangle ABC$ given the conditions about the incenter $K$ and the midpoints, follow these steps:

Given:
- $K$ is the incenter of triangle $\triangle ABC$.
- $C_1$ and $B_1$ are the midpoints of $AB$ and $AC$, respectively.
- $C_1K$ intersects $AC$ at $B_2$.
- $B_1K$ intersects $AB$ at $C_2$.
- The area of triangle $\triangle AB_2C_2$ is equal to the area of triangle $\triangle ABC$.

### Analysis

1. Centroid Property: For median intersecting points of triangles to maintain equal area property, $\triangle AB_2C_2$ being equal in area to $\triangle ABC$ implies reflective symmetry or a special angle configuration.

2. Equal Area Condition:

Since the area of $\triangle AB_2C_2$ is equal to $\triangle ABC$, this condition largely depends on the special properties of angles or symmetries involving the incenter and equal areas.

3. Determine Configuration:

We need to analyze if a special angle or type of triangle would simplify this configuration. If the triangle is equilateral, given it has special symmetry properties, midpoints and intersecting lines describe equal distance and alignment features that could fulfill the condition.

4. Assumption of Equilateral Triangle:

Assume $\triangle ABC$ is equilateral with each angle $60^{\circ}$:
- Here, the incenter coincides with the centroid and orthocenter.
- The lines $C_1K$ and $B_1K$, intersecting at points on sides $AC$ and $AB$, will ensure that such equal area property holds due to symmetry and uniform distance/angle division.

5. Verification:

Most critical angles like $\angle CAB$ in an equilateral triangle are $60^{\circ}$.
The condition comparing the area of triangles $\triangle AB_2C_2$ and $\triangle ABC$ satisfies due to symmetrical bisection of sides by midpoints and equal division through incenter alignment.

Thus, through analysis with considerations of triangle properties, the measure of $\angle CAB$ in $\triangle ABC$ is:
$$\boxed{60^{\circ}}$$