my solution: LET M,N be the midpoint of arc AB,AD first it is easy to prove tangent at A,T,MN are concurrent and $\frac{MK}{NK}=(\frac{AM}{AN})^2$ IN triangle CMN using menelus theorem we have $\frac{CI1}{I1N}*\frac{MI2}{I2C}=\frac{\angle{sinACM}}{\angle{sinACN}}*\frac{AN}{AM}=\frac{MK}{NK}$ so K,I1,I2 are collinear

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duanby wrote: some detail for tangent at A,T,MN are concurrent let L be the midpoint of arcBAD,I be the incenter of BAD, then L,I,T collinear, and by angle chasing done. excuse me, please. can you explain how can we use that L,I,T are collinear? thank you

duanby wrote: my solution: LET M,N be the midpoint of arc AB,AD first it is easy to prove tangent at A,T,MN are concurrent I think that geogebra disproves this. *I didn't see that were the arcs.My fault.

OK,in detail only need to prove K is the circumcenter of AIT ,because MN is the perpendicular bistor of AI TO PROVE $\angle{AIT}+1/2\angle{AKT}=180$ THAT IS $\angle{AIT}-\angle{ALI}=90$ by $AI\perp LA$ done

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Dear Mathlinkers, how is the cyclic quadrilateral ABCD in order to have two equal incircle as the problem? Other way 1. ABC a triangle 2. (O) the circumcirle of ABC 3. How to choose D on (O) in order to have.... Sincerely Jean-Louis

Dear Mathlinkers, any ideas? Sincerely Jean-Louis

hello! my idea is we know MC-MA/NC-NA=NA/MA M,N fixed So MC=$\alpha$NC+$\beta$,where $\alpha$,$\beta$are constant and MN^2=MC^2+NC^2-kMC*NC k=2cosMCN fixed so it's a quadratic equation wrt MC solve it we know it can be done by ruler and compasses

I am very sorry,but can you explain this step? duanby wrote: $\frac{CI1}{I1N}*\frac{MI2}{I2C}=\frac{\angle{sinACM}}{\angle{sinACN}}*\frac{AN}{AM}=\frac{MK}{NK}$ so K,I1,I2 are collinear

Notice that $w'$ is the A-mixtillinear incircle of $\triangle ABD$. Let it be tangent to $AB$ at $R$ and $AD$ at $S$. Let $I$ be the incentre of $\triangle ABD \implies I$ is the midpoint of $RS$. Note that this is a property of the mixtillinear incircles. Let $TR \cap w = M_1, TS \cap w = M_2$, we know that $M_1, M_2$ are midpoints of arcs $AB, AD$ so $M_1, I, D$ and $M_2, I, B$ are collinear. Now, the homothety at $T$ sending $w' \to w$ sends $RS \to M_1M_2$ i.e. $RS \parallel M_1M_2$. So, the pencil $M_1(M_2, I; R, S)$ is harmonic, and intersecting it with $TM_2 \implies (M_2, M_1D \cap TM_2; T, S)$ is harmonic. We have $(M_2, M_1D \cap TM_2; T, S) \stackrel{D}{=} (M_2, M_1; T, A)$, so $TM_1AM_2$ is an harmonic quadrilateral so $K, M_1, M_2$ are collinear. Now, applying Menelaus on $\triangle CM_1M_2$, we need $KM_1/KM_2 \cdot M_2I_1/I_1C \cdot CI_2/I_2M_1 = 1$. Let $X$ be the foot of the altitude of $I_1$ to $AC$ and $Y$ be the foot of the altitude from $I_2$ to $AC$, and note by the condition $IX = IY$. We have $CI_2 = IX/\sin M_1CA$ and $CI_1 = IY/\sin ACM_2$. Also, It is well-known that $M_1$ is the centre of the circle passing through $BI_2A$ so $M_1I_2 = M_1A$ and similarly $M_2I_1 = M_2A$ and so: $\dfrac{KM_1}{KM_2} \cdot \dfrac{M_2I_1}{I_1C} \cdot \dfrac{CI_2}{I_2M_1} = \dfrac{KM_1}{KM_2} \cdot \dfrac{\sin ACM_2}{\sin M_1CA} \cdot \dfrac{M_2A}{M_1A}$. Now note that $\dfrac{M_2A}{M_1A} = \dfrac{KM_2}{KA}$. So: $\dfrac{KM_1}{KM_2} \cdot \dfrac{\sin ACM_2}{\sin M_1CA} \cdot \dfrac{M_2A}{M_1A} = \dfrac{KM_1}{KM_2}\left(\dfrac{M_2A}{M_1A}\right)^2$ by Sine Law. And $(M_1A/M_2A)^2 = (KM_2)^2/KA^2 = (KM_2)^2/(KM_1 \cdot KM_2) = KM_2/KM_1$ by POP. So, $KM_1/KM_2 \cdot M_2I_1/I_1C \cdot CI_2/I_2M_1 = KM_1/KM_2 \cdot KM_2/KM_1 = 1$ as desired $\blacksquare$.