1. Given that triangle \(ABC\) is isosceles with \(AB = AC\), the angle bisectors of \(\angle ABC\) and \(\angle ACB\) meet at \(I\), the incenter of the triangle. We are given that \(\angle CIA = 130^\circ\).

2. Since \(I\) is the incenter, it lies on the angle bisector of \(\angle BAC\). Let \(\angle BAC = 2\alpha\). Therefore, \(\angle BAI = \alpha\) and \(\angle CAI = \alpha\).

3. Let \(\angle ABC = \angle ACB = \beta\). Since \(AB = AC\), \(\triangle ABC\) is isosceles, and thus \(\angle ABC = \angle ACB\).

4. The sum of the angles in \(\triangle ABC\) is \(180^\circ\):
   \[
   \angle BAC + \angle ABC + \angle ACB = 180^\circ
   \]
   Substituting the known angles:
   \[
   2\alpha + 2\beta = 180^\circ \implies \alpha + \beta = 90^\circ
   \]

5. Since \(I\) is the incenter, \(\angle BIC = 90^\circ + \frac{\angle BAC}{2}\):
   \[
   \angle BIC = 90^\circ + \alpha
   \]

6. We are given that \(\angle CIA = 130^\circ\). Note that \(\angle BIC\) can be expressed as:
   \[
   \angle BIC = \angle BIA + \angle AIC = 2\alpha + 2\beta - 180^\circ
   \]

7. Since \(\angle BIC = 90^\circ + \alpha\), we equate this to the given angle:
   \[
   90^\circ + \alpha = 130^\circ
   \]
   Solving for \(\alpha\):
   \[
   \alpha = 130^\circ - 90^\circ = 40^\circ
   \]

8. Therefore, the measure of \(\angle CAB\) is:
   \[
   \angle CAB = 2\alpha = 2 \times 40^\circ = 80^\circ
   \]

The final answer is \(\boxed{80^\circ}\)
Triangle  $ABC$  is isosceles with  $AB=AC$ . The bisectors of angles  $ABC$  and  $ACB$  meet at  $I$ . If the measure of angle  $CIA$  is  $130^\circ$ , compute the measure of angle  $CAB$ .

*Proposed by Connor Gordon*
{'source': 'aops_forum', 'messages': [{'content': 'Triangle  $ABC$  is isosceles with  $AB=AC$ . The bisectors of angles  $ABC$  and  $ACB$  meet at  $I$ . If the measure of angle  $CIA$  is  $130^\\circ$ , compute the measure of angle  $CAB$ .\n\n*Proposed by Connor Gordon*', 'role': 'user'}, {'content': '1. Given that triangle \\(ABC\\) is isosceles with \\(AB = AC\\), the angle bisectors of \\(\\angle ABC\\) and \\(\\angle ACB\\) meet at \\(I\\), the incenter of the triangle. We are given that \\(\\angle CIA = 130^\\circ\\).\n\n2. Since \\(I\\) is the incenter, it lies on the angle bisector of \\(\\angle BAC\\). Let \\(\\angle BAC = 2\\alpha\\). Therefore, \\(\\angle BAI = \\alpha\\) and \\(\\angle CAI = \\alpha\\).\n\n3. Let \\(\\angle ABC = \\angle ACB = \\beta\\). Since \\(AB = AC\\), \\(\\triangle ABC\\) is isosceles, and thus \\(\\angle ABC = \\angle ACB\\).\n\n4. The sum of the angles in \\(\\triangle ABC\\) is \\(180^\\circ\\):\n   \\[\n   \\angle BAC + \\angle ABC + \\angle ACB = 180^\\circ\n   \\]\n   Substituting the known angles:\n   \\[\n   2\\alpha + 2\\beta = 180^\\circ \\implies \\alpha + \\beta = 90^\\circ\n   \\]\n\n5. Since \\(I\\) is the incenter, \\(\\angle BIC = 90^\\circ + \\frac{\\angle BAC}{2}\\):\n   \\[\n   \\angle BIC = 90^\\circ + \\alpha\n   \\]\n\n6. We are given that \\(\\angle CIA = 130^\\circ\\). Note that \\(\\angle BIC\\) can be expressed as:\n   \\[\n   \\angle BIC = \\angle BIA + \\angle AIC = 2\\alpha + 2\\beta - 180^\\circ\n   \\]\n\n7. Since \\(\\angle BIC = 90^\\circ + \\alpha\\), we equate this to the given angle:\n   \\[\n   90^\\circ + \\alpha = 130^\\circ\n   \\]\n   Solving for \\(\\alpha\\):\n   \\[\n   \\alpha = 130^\\circ - 90^\\circ = 40^\\circ\n   \\]\n\n8. Therefore, the measure of \\(\\angle CAB\\) is:\n   \\[\n   \\angle CAB = 2\\alpha = 2 \\times 40^\\circ = 80^\\circ\n   \\]\n\nThe final answer is \\(\\boxed{80^\\circ}\\)', 'role': 'assistant'}]}