Can it be done without trig?

Let $R$ be reflection of $B$ with respect to $AD$. Since $BR$ is perpendicular to $AD$, we get $B, A , R$ are collinear. Thus, $180-\angle DCB=\angle DBA = \angle DBR = \angle DRB $, so $D,C,B,R $ are concyclic.$\angle CAB = 2\angle CDB = 2\angle CRB$, therefore $\angle ACR= \angle CRA$, it means $CA=AR=AB$ and $\angle BCR=90$.Finally, $\angle RBD= \angle RCD = \angle BCD - 90$, however, $\angle RBD + \angle BCD= 180$ ,so $\angle ABD=45$

45 сршйфвш

My solution is quite simple B,C,D are on circumference of circle A. Therefore triangle ABD is right-angle isosceles triangle. Angle ABD = 45.

using trig it is easy

Quote: B,C,D are on circumference of circle A how

Extend DC to P such that BP is perpendicular to CD. Join AP to intersect BC at Q. $ABPD$ is cyclic. So, $ABD$ = $APD$ or, $BCP$ = $APD$ Thus, $QC$ = $QP$. Similar angle chasing gives, $CBP$ = $APB$ Hence, $QB$ = $QP$. Thus, QC = QP = QB, or, Q is the circumcenter of BPC. But, Q lies in the bisector of BAC. Thus, AP is the perpendicular bisector of BC. So, $\Delta BCP$ is isosceles. Hence, $ABD$ = $BCP$ = $45$°

Let $B'$ be the reflection of $B$ in $A$.From the above angle conditions, $\angle AB'S+\angle BCD=180^\circ$.So,$B'BCD$ is cyclic.Now,$\angle BDC=\angle BB'C$.Let $C'$ be the reflection of $C$ in $A$,hence $\angle BB'C=\angle B'BC'$,by angle chase, $\angle BC'C=\angle B'CA$,so $\bigtriangleup B'AC$ is $A$ isosceles.So,$AB=AC=AB'$,hence $A$ is the circumcenter of $\bigtriangleup BB'C$,so $\angle BCB'=90^\circ$,hence $\angle BDB'=90^\circ$.Now,$\bigtriangleup BDB'$ is right angled and $DA \perp BB'$,by angle chase we have $\angle ABD=45^\circ$.

fighter wrote: Quote: B,C,D are on circumference of circle A how If you reflect $B$ and $C$ across $AD$ you have cyclic quadrilaterals $BDC'B'$ and $BCDB'$ as $\angle B'BD+\angle B'C'D=\angle ABD+\angle BCD=180^\circ$. So $BCDC'B'$ is a cyclic pentagon. Furthermore, $A$ lies on the perpendicular bisector of $BB'$, so the circumcircle of $BDB'$ lies on $AD$. As $\angle B'AB=180^\circ$, $A$ is in fact the circumcenter. So $B,C,D$ are on the circumference of circle $A$

Very easy ... 45°