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.