The bisectors of the angle C of the triangle BAC and the outer angle B intersect at point D

The bisectors of the angle C of the triangle BAC and the outer angle B intersect at point D; the angle ABD is 55 degrees; the angle BCD is 40 degrees, find the degree measure CAB

Consider a triangle ABC.

CD is the bisector of BCA and BD is the bisector of the outer corner EBA.

By the condition of the problem, we have that the angles:

BCD = 40 °, ABD = 55 °.

Since BD is the bisector of the angle EBA, then ABD = EBD.

Since CD is the bisector of the angle BCA, then BCD = DCA.

Then we have:

BCA = BCD + DCA = 2 * BCD = 2 * 40 ° = 80 °,

EBA = ABD + EBD = 2 * ABD = 2 * 55 ° = 110 °.

Hence, ABC = 180 ° – EBA = 180 ° – 110 ° = 70 °.

Hence,

CAB = 180 ° – ABC – BCA = 180 ° – 70 ° – 80 ° = 30 °.

Answer: CAB angle = 30 °.