Tangents $ l_1$ and $ l_2$ common to circles $ c_1$ and $ c_2$ intersect at point $ P$, whereby tangent points remain to different sides from $ P$ on both tangent lines. Through some point $ T$, tangents $ p_1$ and $ p_2$ to circle $ c_1$ and tangents $ p_3$ and $ p_4$ to circle $ c_2$ are drawn. The intersection points of $ l_1$ with lines $ p_1, p_2, p_3, p_4$ are $ A_1, B_1, C_1, D_1$, respectively, whereby the order of points on $ l_1$ is: $ A_1, B_1, P, C_1, D_1$. Analogously, the intersection points of $ l_2$ with lines $ p_1, p_2, p_3, p_4$ are $ A_2, B_2, C_2, D_2$, respectively. Prove that if both quadrangles $ A_1A_2D_1D_2$ and $ B_1B_2C_1C_2$ are cyclic then radii of $ c_1$ and $ c_2$ are equal.
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Tangents $ l_1$ and $ l_2$ common to circles $ c_1$ and $ c_2$ intersect at point $ P$, whereby tangent points remain to different sides from $ P$ on both tangent lines. Through some point $ T$, tangents $ p_1$ and $ p_2$ to circle $ c_1$ and tangents $ p_3$ and $ p_4$ to circle $ c_2$ are drawn. The intersection points of $ l_1$ with lines $ p_1, p_2, p_3, p_4$ are $ A_1, B_1, C_1, D_1$, respectively, whereby the order of points on $ l_1$ is: $ A_1, B_1, P, C_1, D_1$. Analogously, the intersection points of $ l_2$ with lines $ p_1, p_2, p_3, p_4$ are $ A_2, B_2, C_2, D_2$, respectively. Prove that if both quadrangles $ A_1A_2D_1D_2$ and $ B_1B_2C_1C_2$ are cyclic then radii of $ c_1$ and $ c_2$ are equal.
Let the incircle of $ \triangle A_2PA_1$ have radius $ r$ and the incircle of $ \triangle D_1PD_2$ have radius $ R$. Notice that since $ B_2C_1C_2B_1$ and $ A_1A_2D_1D_2$ are cyclic, we have that $ \angle TA_1B_1 = \angle TD_2C_2$ and $ \angle TB_1P = \angle TC_2P$. Thus, $ \angle TB_1P - \angle TA_1B_1 = \angle TC_2P - \angle TD_2C_2$. It follows that $ \angle A_1TB_1 = \angle C_2TD_2$, so $ \triangle TB_1A_1\sim \triangle TC_2D_2$. Furthermore, $ \triangle A_2PA_1\sim \triangle D_1PD_2$. Thus, their incircles are in porportion, so $ \frac {A_1P}{D_2P} = \frac {r}{R}$. Yet, $ \triangle B_2PB_1\sim \triangle C_1PC_2$, so their excircles are in porportion, so $ \frac {PB_1}{PC_2} = \frac {r}{R} = \frac {A_1P}{D_2P}$. This, combined with the fact that $ \triangle TB_1A_1\sim \triangle TC_2D_2$ implies that $ \triangle TPA_1\sim \triangle TPD_2$, so $ \angle A_2TP = \angle D_1TP$. Also, \[ \angle TPA_2 = \angle TPA_1 - \angle A_2PA_1 = \angle TPD_2 - \angle D_1PD_2 = \angle TPD_1\] Furthermore, $ TP = TP$, so $ \triangle TPA_2\cong \triangle TPD_1$. It follows that $ TA_2 = TD_1$. Since $ A_1A_2D_1D_2$ is cyclic, we conclude that $ A_1A_2 = D_1D_2$, so $ \triangle A_1A_2P\cong \triangle D_2D_1P$. Thus, $ r = R$, as desired.
