Let $I$ be the circumcenter of $\triangle ACE.$ Clearly $ABCI,$ $CDEI,$ $EFAI$ are kites $\Longrightarrow$ $IB,ID,IF$ bisect $\angle ABC,$ $\angle CDE,$ $\angle EFA.$ Moreover, $\angle IAF=\angle IEF$ $\Longrightarrow$ $\angle IAB=\angle FAB-\angle IAF=\angle DEF-\angle IEF=\angle IED,$ but $\angle IAB=\angle ICB$ and $\angle IED=\angle ICD$ $\Longrightarrow$ $\angle ICD=\angle ICB,$ i.e. $IC$ bisects $\angle BCD.$ Similarly, $IA,IE$ bisect $\angle FAB,$ $\angle DEF$ $\Longrightarrow$ $ABCDEF$ has an incircle with center $I,$ thus by Brianchon theorem $AD,BE,CF$ concur.