The idea of the problem is how to use the incenters here. If you notice that $\measuredangle{ABC}= \measuredangle{ABD}$ and triangles $ABC$ and $DBA$ are similar, you are in the right track. Proof It is first easy to see that $\measuredangle{BAC}= \measuredangle{BDA}, \measuredangle{ACB}=\measuredangle{BAD}$. Hence $\measuredangle{ABC}= \measuredangle{ABD}=\beta$ and $ABC$ and $DBA$ are similar. It follows that $\measuredangle{I_{1}BI_{2}}= \beta$ and $\frac{BI_{1}}{BI_{2}}= \frac{BC}{BA}$. Thus, $I_{1}BI_{2}, ABC, DBA$ are similar. Let $M$ be the intersection of $BI_{1}$ and $AC$. Then $\frac{BE}{BM}= \frac{BI_{1}}{BA}$. Since $BM = \frac{2BA.BC.cos \frac{\beta}{2}}{BA+BC}$ and $2.BI_{1}cos \frac{\beta}{2}= BC+BA-AC$, we obtain$\frac{BE}{AB}= \frac{BA+BC-CA}{BC+BA}$ Thus $\frac{AE}{AB}= \frac{AC}{BA+BC}$. Therefore, $\frac{1}{AE}= (\frac{BC+BA}{AB}) . \frac{1}{AC}= \frac{1}{AC}+\frac{1}{AD}$.

Lemma: $ H$ is orthocenter of a $ \triangle ABC,$ the A-altitude $ AH$ cuts the triangle circumcircle $ (O)$ again at $ X.$ $ AH_1, AH_2$ are reflections of $ AH$ in $ AB, AC.$ Parallels through $ H$ to $ AH_2, AH_1$ cut $ AH_1, AH_2$ at $ K_1, K_2,$ forming a parallelogram $ AK_1HK_2.$ Then $ AK_1 + AK_2 = AX.$ Let $ (O_1),\ X_1 \in (O_1)$ be reflections of $ (O),\ X \in (O)$ in $ AB.$ $ \angle AHK_1 = \angle HAK_2 = 180^\circ - 2 \angle C$ $ \Longrightarrow$ $ \angle XHK_1 = 2 \angle C.$ As $ \angle BHX = \angle HXB = \angle C,$ $ HB$ bisects the angle $ \angle XHK_1.$ Obviously, $ AB$ bisects the angle $ \angle HAK_1.$ Then $ B$ is the A-excenter of the $ \triangle AHK_1$ and $ K_1B$ bisects the $ \angle X_1K_1H.$ It follows that $ X_1BHK_1$ with $ X_1B = XB = HB$ is a kite and $ K_1X_1 = HK_1.$ In conclusion, $ AK_1 + AK_2 = AK_1 + HK_1 = AK_1 + K_1X_1 = AX_1 = AX.$ Problem: $ (O), (P)$ are the 2 circles intersecting at $ A, B.$ $ (O), (P)$ cut the perpendicular bisector $ OP$ of $ AB$ at $ M, N$ on the opposite sides of $ AB$ than $ C, D.$ Let $ AB, AC, AD$ cut $ OP$ at $ K, U, V.$ Then $ (O), (P)$ are A-Apollonius circles of the right $ \triangle AKV, \triangle AKU$ $ \Longrightarrow$ $ AM, AN$ bisect the angles $ \angle DAB, \angle CAB.$ Circles $ (M), (N)$ with radii $ MA, NA$ cut internal bisectors $ AN, AM$ of the angles $ \angle CAB, \angle DAB$ at the incenters $ I, J$ of the $ \triangle ABC, \triangle ABD.$ The circles $ (M), (N)$ are centered on $ AJ, AI,$ perpendicular to $ AJ, AI.$ Inversion with center $ A$ and power $ AB^2$ takes $ (O), (P)$ into parallels $ o \parallel AD, p \parallel AC$ through $ B$ to their tangents $ AD, AC.$ If $ C', D'$ are inversion images of $ C, D,$ then $ AC'BD'$ is a parallelogram, such that lines $ AC', AD'$ are reflections of the line $ AB$ in $ AI, AJ.$ Lines $ AI, AJ$ go to themselves and circles $ (M), (N)$ to perpendiculars $ m \perp AJ, n \perp AI$ through $ B,$ intersecting $ AI, AJ$ at the inversion images $ I', J'$ of $ I, J.$ Thus $ B$ is orthocenter of $ \triangle AI'J'.$ Line $ IJ$ goes to its circumcircle $ (Q)$. $ E$ goes to intersection $ E'$ other than $ A$ of the 3rd antitude $ AB$ of the $ \triangle AI'J'$ with its circumcircle $ (Q).$ By the above lemma, $ AE' = AC' + AD'$ $ \Longrightarrow$ $ \frac {AB^2}{AE} = \frac {AB^2}{AC} + \frac {AB^2}{AD}.$ EDIT: Hey, this was reply to a problem posted by shobber today and he did not mention any $ \omega_2$ !!!

Before reading this proof, one has to ensure that he is in a good mental state to read an unexpectedly large solution. My proof runs as follows. Firstly note that $AC$ and $AD$ are tangents to $\omega_2$ and $\omega_1,$ respectively. Therefore we obtain \[\begin{aligned}&\angle BCA+\angle CAB=\angle BAD+\angle ADB;\\&\implies \angle ABC=\angle ABD\implies \angle ABI_1=\angle ABI2.\end{aligned}\] Denote $G=BI_1\cap AC, F=BI_2\cap AD, X=BI_1\cap \omega_1, Y=BI_2\cap \omega_2.$ Since $\pi-\angle CAY=\angle YBA=\frac{\angle XBY}{2}=\angle XAG,$ therefore $X,A,Y$ are collinear. Now, note that $\begin{aligned}\angle GAF&=\pi-(\angle XAC+\angle YAD)\\&=\pi-2\cdot\frac{\angle XBY}{2}=\pi-\angle XBY;\end{aligned}$ So $AGBF$ is a cyclic quadrilateral. Since $\angle GBA=\angle FBA,$ therefore $\angle GAF=\angle GFA\implies AG=AF.$ Again, note that using this in accordance with the angle-bisector theorem, we obtain $\frac{AG}{AB}=\frac{AF}{AB}\implies \frac{BI_1}{I_1G}=\frac{BI_2}{I_2F},$ implying that $I_1I_2\parallel GF.$ Since $\angle GFB=\angle CAB=\angle AYB,$ therefore $XY\parallel GF,$ so that $I_1I_2\parallel GF\parallel XY.$ Using this we obtain that $\begin{aligned}\angle I_1EA=\angle EAY&=\angle DAB+\angle DAY=\angle ACB+\angle DAY\\&=\angle ACB+\angle DBY=\angle ACB+\angle GBC\\&=\angle I_1GA.\end{aligned}$ Therefore $\triangle AGI_1\cong\triangle AEI_1\implies AG=AE=AF.$ On the other hand, we have $\frac{AF}{FD}=\frac{CB}{AB};$ or, $\frac{AF}{AD}=\frac{AB}{BD+AB}.$ By similar means we have its analogue $\frac{AG}{GC}=\frac{AB}{CB+AB},$ so the relation that we are required to prove is equivalent to with $\frac{AG}{AC}+\frac{AF}{AD}=1, \iff \frac{AB}{CB+AB}+\frac{AB}{BD+AB};$ Which is equivalent to $AB^2=BC\cdot BD.$ This time, note that $\triangle BCG\sim\triangle BAX$(Tuberculosis!)$\implies \boxed{BG\cdot BX=AB\cdot BC}.$ Analogously we obtain $\boxed{BF\cdot BY=AB\cdot BD}.$ Multiplying the last two relations leads to $\boxed{AB^2\cdot BC\cdot BD=BG\cdot BX\cdot BF\cdot BY} \ \ \ \ \ (*).$ Now, note that $\left.\begin{array}{ccc}\angle GBA&=\angle YBA\\ \angle GAB&=\angle AYB\end{array}\right\}\implies \triangle AGB\sim\triangle AYB;$ So that $\boxed{AB^2=BY\cdot BG}.$ From $\triangle ABX\sim\triangle ABF$ we obtain its analogue $\boxed{AB^2=BX\cdot BF}.$ Multiplying the last two relations together we get $\boxed{BX\cdot BF\cdot BY\cdot BG=AB^4}. \ \ \ \ \ (**).$ Substituting the RHS in $(*)$ with RHS of $(**)$ we get \[AB^2\cdot BC\cdot BD=AB^4\implies AB^2=BC\cdot BD;\] As required! We are done.$\Box$

Note

Attachments:

That is my proof. Let $O_{1}$,$O_{2}$ is the center of circumcircle of $ABC$,$ABD$. $O_{1}O_{2}$ meets $AI_{1}$,$AI_{2}$ at $M$,$N$. $BI_{1}$ meets $AC$ at $Q$ and $BI_{2}$ meets $AD$ at $P$. Easy to see that: $A$,$M$,$I_{1}$ is collinear and $A$,$N$,$I_{2}$ is collinear. Have $\angle MI_{1}N=\angle MAN=\angle MI_{2}N$ hence $MNI_{1}I_{2}$ is inscribed. then $\angle AI_{1}E=\angle ANM=\frac{\pi}{2}-\angle ACB=\angle AI_{1}Q$ hence $AE=AQ$ similar $AE=AP$ Then have $\frac{AE}{AD}=\frac{AB}{AB+BD} and \frac{AE=AQ}{AC}= \frac{AB}{AB+BC}$(*) Note that $BC.BD=AB^{2}$ (**) When (*).(**) we have QED.

Mine is an algebraic solution: clearly triangles $ADB$ and $CAB$ are similar. let $CA=r.AD$ where $r<1$ (assuming that CAB is smaller than ADB) and $AD=x.AB$ now we normalize $AB=1$ then we get $AD=x$ $BD=1/r$ $BC=r$ $AC=rx$ so we have $\frac{2}{\frac{1}{AC}+\frac{1}{AB}}=\frac{2rx}{1+r}$ thus we need to show that $2AE(1+r)=2rx$. drop perpendiculars from $I_1$ and $I_2$ on $AB$ at $P$ & $Q$ respectively. so, $2PB=AB+BC-AC$ and $2PQ=AB+BD-AD$ thus, $PQ=(1-r)(1+\frac{1}{r}+x)$ also $I_1PE$~$I_2QE$ with $PE=r.QE$ thus $2QE=(\frac{1-r}{1+r})(1+\frac{1}{r}+x)$ also $2AQ=AB+AD-BD=1+x-\frac{1}{r}$ and thus $2AE(1+r)=2rx$ as desired.