Does not seem true...

oneplusone wrote: Does not seem true... Sorry, it should be $\angle BAC=30^{\circ}$.

Are you sure it works?

It's false.

We state R as the reflection of B across AC .Thus let M be the intersection point of CA diagonal and BR. We notice the right angled triangle ABM and triangle ARM are congruent, and so we have angle ABR = ARB and angle BAM = RAM.Now we have triangle ABR to be equilateral, so AB = BR = AR. Also notice that we can circumscribe a circle with B as the center and radius AB. Thus it passes through points A, R and D (We have this since 2*angle ADR = RBA or AB = BR as radius). Since angle ABR = 60° ,then angle ADR = 30° and : angle RDC = RDA + ADC = 30° + 150° = 180° then points D, C and R are collinear. Now we can easily see the congruence of triangle BCM and triangle RCM which leads to angle BCM=ACD or angle BCA = ACD.

Steve12345 wrote: We state R as the reflection of B across AC .Thus let M be the intersection point of CA diagonal and BR. We notice the right angled triangle ABM and triangle ARM are congruent, and so we have angle ABR = ARB and angle BAM = RAM.Now we have triangle ABR to be equilateral, so AB = BR = AR. Also notice that we can circumscribe a circle with B as the center and radius AB. Thus it passes through points A, R and D (We have this since 2*angle ADR = RBA or AB = BR as radius). Since angle ABR = 60° ,then angle ADR = 30° and : angle RDC = RDA + ADC = 30° + 150° = 180° then points D, C and R are collinear. Now we can easily see the congruence of triangle BCM and triangle RCM which leads to angle BCM=ACD or angle BCA = ACD. Your proof used the fact that $R$ lies onto $(CD$, which we need to prove, but it did not use $CD=AB$, hence it is wrong! It seems we need more data to solve this problem. $D$ lies onto a circle with center $X$, reflection of $C$ across $AB$, $\triangle ACX$ being equilateral, and radius $XA$ and we do not usually get $CD=AB$ when $\angle ACD=\angle ACB$ Best regards, sunken rock

I constructed it and got the needed AB = CD even though i didnt use it.Weird.

If $AB=AD$, then the problem would be consistent. In this case, taking the symmetry of point $B$ wrt the diagonal $AC$, we denote as $B'$. Then, the circle centered at $B$ with radius $AB$ would pass through points $B'$ and $D$. So we get $\angle ADB'=30^\circ$ , and points $B'$, $D,$ $C$ would be collinear.