There is a triangle ABC with point D at the center of AB.

Angle CDA = 45 degree

Angle DCB = 15 degree

Need to find the angle CAD.

I figured that since CD is median that can divide the triangle in two smaller triangle of equal areas, I can use that property.

Area of Triangle CAD = CA*sin(x)*AD/2

Area of Triangle ABC= AB*sin(30)*BC/2

Area of Triangle DCB= DC*sin(15)*CB/2

Area of Triangle ABC = Area of Triangle CAD + Area of Triangle DCB

This approach didn't help. Any idea how to proceed?