Austrian Mathematical Olympiad

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.