In triangle ABC, the bisectors of the outer angles at vertices A and B intersect at point D.

In triangle ABC, the bisectors of the outer angles at vertices A and B intersect at point D. Find the angle BDA if the BCA angle is 32 degrees.

Let the angle DBA = X0, and the angle DAB = Y0, then the angle ABK = 2 * X0, the angle BAM = 2 * Y0.

In triangle ABC, angle ABC = (180 – 2 * X), angle BAC = (180 – 2 * Y).

Then in the triangle ABC, 180 = (180 – 2 * X) + (180 – 2 * Y) + 32.

2 * X + 2 * Y = 180 + 32 = 212.

X + Y = 106.

In a triangle ABD (X + Y) + BDA = 180.

DBA = 180 – (X + Y) = 180 – 106 = 74.

Answer: The BDA angle is 74.