In the isosceles triangle $AUB$, we have
$$\angle BAU = \angle UBA = 60^\circ - 45^\circ = 15^\circ,$$
and therefore
$$\angle AUB = 180^\circ - \angle BAU - \angle UBA = 150^\circ.$$
The inscribed angle theorem implies
$$\angle BCA = \frac{1}{2} \angle BUA = 75^\circ,$$
and therefore
$$\angle BAC = 180^\circ - 60^\circ - 75^\circ = 45^\circ.$$
We can finally compute the two angles of interest:
$$\angle UAD = \angle BAD - \angle BAU = \angle BAC - \angle BAU = 45^\circ - 15^\circ = 30^\circ$$
$$\angle DUA = 180^\circ - \angle AUB = 180^\circ - 150^\circ = 30^\circ$$
Therefore, the triangle $AUD$ is isosceles with apex $D$ and we have $AD = DU$ as desired.