The answer is $15-75-90.$ Let $\angle{ABD}=\alpha$ then $\angle{BAD} = 30-\alpha$ and $\angle{ACB}=90-\alpha.$ From Law of Sines in $\triangle{ABD}$ we have. $$\dfrac{\sin(150)}{AB} = \dfrac{\sin(30-\alpha}{BD}.$$From Law of Sines in $\triangle{BDC}$ we have $$\dfrac{\sin(30)}{BC} = \dfrac{\sin(90-\alpha)}{BD}.$$Multiplying and using the length condition gives $$\dfrac{1}{4} = \sin(30-\alpha)\sin(90-\alpha) = \sin(30-\alpha)\cos(\alpha) = \dfrac{1}{2}\sin(30) + \dfrac{1}{2} \sin(30-2\alpha).$$This means $\sin(30-2\alpha)=0 \rightarrow \alpha=15.$ It follows that the angles of the triangle are $15-75-90$ as desired.

MathLuis wrote: On $\triangle ABC$ if there existed a point $D$ in $AC$ such that $\angle CBD=\angle ABD+60$ and $\angle BDC=30$ and $AB \cdot BC=BD^2$, then find the angles inside the triangle $\triangle ABC$ Let Construct $O$ as Circumcenter of $\triangle BDC$ then observe $OB=OC=OD=BC$ and also $\angle DBO=\angle BDO=\angle ABD$ Now observe $\triangle ABD\sim \triangle BDO$ Because by SAS Similarity we have $\frac{AB}{BD}=\frac{BD}{BC}=\frac{BD}{OD}$ and $\angle ABD=\angle BDO$. So $\boxed{\angle ABD=\angle BAC}$ and hence angles $B, C, A$ respectively are $90^\circ, 75^\circ, 15^\circ$

Tense moments when u get no ideas. Let $P$ in $AC$ such that $BD,BP$ are isogonal w.r.t. $\angle ABC$ then $\angle CBD=\angle PBC+60$ gives that $\angle DBP=60$ and that gives $BP \perp AC$ and now since we have a 30-60-90 config on $\triangle BDP$ we get that $BD=2BP$ and now by $\sqrt{ac}$ inversion we have that $P'$ is the B-antipode in $(ABC)$, now let $R$ the radius of $(ABC)$ then we get that $BD \cdot R=AB \cdot BC=BD^2$ which means that $BD=R$ but since $BD$ is isogonal to the B-altitude we have that $BD$ passes through the center of $(ABC)$ meaning that $D$ is the center of $(ABC)$ and since it lies in $AC$ we get that $\angle ABC=90$ so $\angle ABD=\angle PBC=15$ and both mean that $\angle BAC=15$ and $\angle BCA=75$, hence $\triangle ABC$ has angles 15-75-90 thus we are done

Cono Sur 2011 P5

Let angle ABD=x So, angle CBD=x+60° angle BDC=30° so, angle BAC=30°-x angle ACB=90°-x By, sine rule in ∆BDC, BD/BC=sin(90°-x)/sin 30° = 2cosx In ∆ ABD,by sine rule, BD/ AB= sin(30°-x)/sin 30°=2sin(30°-x) Since BD²=AB.AC So, 2cosx=1/2sin(30°-x) 2sin(30°-x)cos x=1/2 sin 30°+ sin(30°-2x)=1/2 sin(30°-2x)= 0 So, 30°-2x=0 x=15° So, the angles are 15°,75°,105°