Consider triangles $O_1CX$ and $O_2DY$. Now applying law of sines we are left to show a fact which arises from law of sines in $O_1AO_2$.

Wow! Congrats to Alireza! I had the opportunity to meet him at the Final Round of the Sharygin Geometry Olympiad in 2016.

Sorry for bumping, but can someone send a figure for this problem? I am not able to make it

Math-wiz wrote: Sorry for bumping, but can someone send a figure for this problem? I am not able to make it

TheDarkPrince wrote: Consider triangles $O_1CX$ and $O_2DY$. Now applying law of sines we are left to show a fact which arises from law of sines in $O_1AO_2$. Could you please give a detailed solution.

Let $B'$ and $A'$ be the antipodes of $B$ and $A$ on circles $\omega_1$ and $\omega_2$. In directed angles: $\angle BCA'=\angle BAA'=\angle BAD=\angle BB'D$ This implies that $B'D \parallel A'C$. Because $A'C \perp AC$ and $B'D \perp BD,$ it follows that $BD \parallel AC$ implying that $$DA=BX$$and $$BC=YA$$. I claim that $\triangle {DAY}$ is similar to $\triangle {XBC}$. Proof: $\angle YDA=\angle BDA=\angle BXC$ by cyclicity $\angle DYA=\angle ACB=\angle XCB$. Hence, $\triangle {DAY}$ is similar to $\triangle {XBC}$ by AAA similarity. Note that the similarity ratio is $1=\frac{DA}{BX}=\frac{YA}{BC}$. From this, we deduce that $$DY=CX$$

Solution. The main idea is to show that $DXCY$ is a parallelogram which implies $CX=DY$. By Basic Angle chasing or Reim's theorem, $DX||YC$. Let $\angle DAX =a$. Then $\angle AO_2C=\pi-2a \implies \angle ABC = \frac \pi 2 +a \implies \angle AO_1B=2a \implies \angle ADB = a$. Hence, $\angle DAX = \angle ADB$ so $XC||DY$ and $DXCY$ is a parallelogram.

I claim that $BCX\cong AYD$. We have that $$\angle BXC=180^\circ-\angle AXB=\angle BDA=\angle ADY$$and $$\angle DYA=180^\circ-\angle AYB=\angle ACB=\angle XCB,$$hence $BCX\sim AYD$. To show that these triangles are congruent, I show that $AY=BC$, which is equivalent showing that $BY\parallel AC$. We have that $$\angle BO_1X=2\angle BAX=2\angle BAC=\angle BO_2C\implies \angle O_1BX=\angle O_1CO_2,$$hence $BX\parallel CO_2$. Also, we have that $$\angle DBX=180^\circ-\angle DAX=\angle CAO_2=\angle ACO_2.$$Hence, by the last two we conclude that $BY\parallel AC$ and we are done.

Let $AO_2 \cap \omega_2=F$ then $FA$ is diameter of $\omega_2$. Let $BO_1 \cap \omega_1=E$ then $BE$ is diameter of $\omega_1$. Now by Reim's we have $DX \parallel YC$ and $DE \parallel FC$ but since $\angle XCF=90=\angle YDE$ we have that $DY \parallel XC$ thus $DXCY$ is a parallelogram which means $DY=CX$ as desired.

Lets show CY = BA = DX. we will prove DY || CX so CYBA and DXAB will be isosceles trapezoids. ∠BO1O2 = ∠BO1A/2 = ∠BDO2 ---> DO1BO2 is cyclic. ∠ACB = ∠AO2B/2 = ∠AO2O1 = ∠DBO1 = ∠YBC ---> DY || XC. we're Done.

$BC \cap \omega_1=B,K,DA \cap \omega_2=L,KA \cap DB=P,DB \cap CL=Q$ Let $\angle BLD=\alpha$ and $\angle DKA=\beta$ We have $\angle KBD=\angle KAD=\angle BLD=\angle BCA=\alpha $ so $DB \parallel AC$ We'll prove that $\frac{DA.DL}{DB}=\frac{CB.CK}{CA}$ We have \[\frac{CA}{CK}=\frac{sin \angle 90-\alpha-\beta}{sin \angle 90-\beta}\]\[\frac{DA}{DB}=\frac{sin \angle \beta}{sin \angle 90- \alpha}\]$\implies \frac{CA}{CK}.\frac{DA}{DB}=\frac{sin \angle 90-\alpha-\beta}{sin \angle 90-\beta}.\frac{sin \angle \beta}{sin \angle 90- \alpha}$ So \[\frac{BC}{DL}=\frac{\frac{QC}{sin \angle \alpha}}{\frac{QL}{sin \angle 90-\alpha-\beta}}=\frac{QC}{QL}.\frac{ sin\angle 90-\alpha-\beta}{sin \angle\alpha}=\frac{DA}{DL}.\frac{ sin\angle 90-\alpha-\beta}{sin \angle\alpha}=\frac{DP}{DB}.\frac{ sin\angle 90-\alpha-\beta}{sin \angle\alpha}=\frac{DP}{KD}.\frac{KD}{DB}.\frac{ sin\angle 90-\alpha-\beta}{sin \angle\alpha}\]\[=\frac{sin \angle \alpha}{\sin \angle 90-\alpha}.\frac{sin \angle \beta}{sin \angle 90-\beta}.\frac{ sin\angle 90-\alpha-\beta}{sin \angle\alpha}=\frac{sin \angle \beta}{sin \angle 90-\alpha }.\frac{sin \angle 90-\alpha-\beta}{sin \angle 90-\beta}=\frac{CA}{CK}.\frac{DA}{DB}\]Which means $BC.CK.DB=CA.DA.DL \iff \frac{DA.DL}{DB}=\frac{CB.CK}{CA}$ as desired.