Since $\angle CBU = 45^\circ$, we find that $\angle BUC=90^\circ$ and $\angle BAC=45^\circ$. $\angle UAD=30^\circ$ and $\angle ADU=180^\circ-\angle UDC=180^\circ-(90^\circ-\angle UCD)=120^\circ$. Thus, $\angle AUD=30^\circ=\angle UAD$. Then, $|AD|=|DU|$, as desired.

What they ask us is equivalent to: $\angle AUD= \angle DAU $ Notice that $\angle ABD=15$ $\implies \angle AUD=30$ With chasing angles $\angle DAU =30.$

Notation : $(XY)$ stands for small arc $XY$ Let $BU$ intersect the circumcircle at $M$ $\angle AUD=(AM)=2\angle ABU =2*15^o=30^o$ $(AB)=\angle BUA=180^o-2\angle ABU=180^o-2*15^o=180^o-30^o=150^o$ $(MC)=2\angle MBC= 2*45^o=90^o$ Due to lemma $\angle UOA= (AB)/2 + (MC)/2=(150^o+90^o)/2 =120^o$ So by angles triangle sum in $UAO,$ $\angle UAO=30^o=\angle AUD \Rightarrow AD=UD$ Lemma: If two chords $XY$ and $ZW$ intersect inside their circle at point $K$, $\angle XKW= ( (YW)+(ZY) ) /2$

$\measuredangle CBU+\measuredangle UBA=\measuredangle CBA\Rightarrow \measuredangle UBA=15$ $\triangle CBU$ isosceles$\Rightarrow \measuredangle UCB=\measuredangle CBU=45$ and $\measuredangle BUC=90$ $\triangle ABU$ isosceles$\Rightarrow \measuredangle BAU=\measuredangle UBA=15$ and $\measuredangle AUB=150$ $\measuredangle AUB+\measuredangle BUC+\measuredangle CUA=360\Rightarrow \measuredangle CUA=120$ $\triangle CAU$ isoscoles$\Rightarrow \measuredangle UAC=\measuredangle ACU=30$ because $\measuredangle UAC+\measuredangle ACU+\measuredangle CUA=180$ (Also note that $\measuredangle UAC=\measuredangle UAD$ because $C;D;A$ is colinear, so $\measuredangle UAD=30$) $\measuredangle DUA+\measuredangle AUB=\measuredangle DUB=180\Rightarrow \measuredangle DUA=30$ $\measuredangle DUA=30=\measuredangle UAD\Rightarrow AD=DU$