Hello, acc100jt!
I think I've got it . . .
Draw radii XA, XB, and XD.
∆XAD is isosceles: ./XAD = /XDA = α
∆XBD is isoscles: ./XDB = /XBD = β . → . /CBX .= .180° - B
Hence: ./D .= ./XDA + /XDB .= .α + β
CAXB is a cyclic quadrilateral; its opposite angles are supplementary.
Since /CBX = 180° - β, then /CAX = β.
Hence: ./A .= ./XAD + /XAC .= .α + β
Therefore: ./A = /D, .CA = CD . → . ∆ACD is isosceles.