Let \[ \ell_3 \parallel \ell_2, A\in \ell_3, AD\cap BC=K, \ell_3\cap BC=Y, \ell_2\cap AD=X, \ell_3\cap CD=L, \ell_2\cap AB=M \] The quadrilateral $CXAB$ are circumscribed, then $MC+AK=CK+AM$. Then $AL+AK=CK+CL$ and the quadrilateral $CDAY$ are circumscribed, hence $\ell_3=\ell_1$.

Fagot you wrote that fagot wrote: The quadrilateral $CXAB$ are circumscribed, then $MC+AK=CK+AM$. . but in my figure this doesnot hold since if I take equivalently this equality I found that a positive number is equal with a negative.Could you please explain or post the figure in which you solve the problem ?Thanks

Another solution: Denote by $K$ the point of intersection of $AD$ and $BC$. The center of $\omega_2$ is the incenter of $DCK$. Center $E'$ of $\omega_1$ is the image of the center $E$ of the excircle of $DCK$ ,corresponding to the edge $DC$, trough o homothety of center $K$, and ratio $KA/KD$. It's easy to check that triangles $KDI$ and $KCE$ are similar. Since $KA/KD=KE'/KE$ the two figures $K,I,A,D$ and $K,C,E,E'$ are similar, hence $\angle KIA\equiv \angle E'CK$. We have that $\angle DAI=\frac{\angle DKI} 2+\angle AIK=\frac{\angle DKI} 2+\angle E'CK \Rightarrow \\ 2\angle DAI=\angle DKI+\angle 2\angle KCE'$. (1) The angle between tangents from a pooint to a circle is twice the angle between a tangent and the line joining that point to the ccenter of that crcle. Hence, the angle between $BK$ and $\ell_2$ is $2\angle KCE'$ and the angle between $KD$ and $\ell_1$ is $2\angle DAI$. From (1) we can get that $\ell_1$ and $\ell_2$ intersect $KC$ at same angles, hence are paralell.

Quote: but in my figure this doesnot hold since if I take equivalently this equality I found that a positive number is equal with a negative.Could you please explain or post the figure in which you solve the problem ?Thanks The segments with signs: if $D\in [AX]$ and $A\in [DK]$ then $AK-CM=CK-AM$, hence $AK-AL=CK-CL$ and $CDAY$ inscribed, e. t. c. Quote: the two figures $K,I,A,D$ and $K,C,E,E'$ are similar the two figures $K,I,A,D$ and $K,C,E',E$ are similar Quote: $\angle DAI={\angle DKI\over 2}+\angle AIK$ $\angle DAI=\angle DKI+\angle AIK={\angle DKC\over 2}+\angle AIK$...... $2\angle DAI=\angle DKC+2\angle KCE'$.

Could someone post a picture :

xirti wrote: Another solution: Denote by $K$ the point of intersection of $AD$ and $BC$. The center of $\omega_{2}$ is the incenter of $DCK$. Center $E'$ of $\omega_{1}$ is the image of the center $E$ of the excircle of $DCK$ ,corresponding to the edge $DC$, trough o homothety of center $K$, and ratio $KA/KD$. It's easy to check that triangles $KDI$ and $KCE$ are similar. Since $KA/KD=KE'/KE$ the two figures $K,I,A,D$ and $K,C,E,E'$ are similar, hence $\angle KIA\equiv \angle E'CK$. We have that $\angle DAI=\frac{\angle DKI}2+\angle AIK=\frac{\angle DKI}2+\angle E'CK \Rightarrow \\ 2\angle DAI=\angle DKI+\angle 2\angle KCE'$. (1) The angle between tangents from a pooint to a circle is twice the angle between a tangent and the line joining that point to the ccenter of that crcle. Hence, the angle between $BK$ and $\ell_{2}$ is $2\angle KCE'$ and the angle between $KD$ and $\ell_{1}$ is $2\angle DAI$. From (1) we can get that $\ell_{1}$ and $\ell_{2}$ intersect $KC$ at same angles, hence are paralell. Sorry, I'm newbie but could you(or someone) explain me why triangle $KDI$ and $KCE$ are similar?

fagot wrote: Let \[\ell_{3}\parallel \ell_{2}, A\in \ell_{3}, AD\cap BC=K, \ell_{3}\cap BC=Y, \ell_{2}\cap AD=X, \ell_{3}\cap CD=L, \ell_{2}\cap AB=M\] The quadrilateral $CXAB$ are circumscribed, then $MC+AK=CK+AM$. Then $AL+AK=CK+CL$ and the quadrilateral $CDAY$ are circumscribed, hence $\ell_{3}=\ell_{1}$. If $\ AD\parallel BC$ then...?

M4RI0 wrote: Could someone post a picture : Here is a picture. Image not found To fagot: Could you or somebody else explain your solution in details on my image?