Let $M_1$ be midpoint of $AC$ and $M_2$ be midpoint of $BD$ and $M_3$ of $FC$ and $M_4$ of $AE$ and $M_6$ be midpoint of $AB$ and $M_5$ be midpoint of $BC$ . it suffices to show $$\angle{M_4M_1M_3}=\angle{M_4M_2M_3}$$ we have $M_1M_3||BA$ and $M_1M4||BC$ then $\angle{M_3M_1C}=\angle{BAC}$ and $\angle{M_4M_1A}=\angle{BCA}$ that implies $$\angle{M_4M_1M_3}=180-\angle{BAC}-\angle{BCA}=\angle{ABC}$$ Claim: $\angle{M_4M_2M_6}=\angle{M_5M_2M_3}$ observe that $M_5M_3=\frac{BF}{2}=\frac{BE}{2}=M_6M_4$ [tangent from same point] and $M_2M_5=\frac{DC}{2}=\frac{AD}{2}=M_6M_2$ and $\angle{M_3M_5B}=180-\angle{FBC}$ and $\angle{BM_5M_2}=\angle{BCD}$ then we have $$\angle{M_3M_5M_2}=\angle{BCD}-(180-\angle{ABC})=\angle{ABC}+\angle{BCD}-180$$ similarly $$\angle{M_4M_6M_2}=180-\angle{BAD}-\angle{ABC}=180-\angle{BAD}-\angle{ADC}=\angle{ABC}+\angle{BCD}-180$$[sum of angles of a quadrilateral is 360] hence $M_3M_5M_2$ and $M_4M_6M_2$ are congruent by $SAS$ congruence. so $\angle{M_5M_2M_3}=\angle{M_6M_2M_4}$ then its easy to see $\angle{M_4M_2M_3}=\angle{M_6M_2M_5}-\angle{M_5M_2M_3}+\angle{M_6M_2M_4}=\angle{M_6M_2M_5}=\angle{BM_2M_5}+\angle{BM_2M_6}$ as $M_2M_6||AD$ and $M_2M_5||BC$ so $\angle{BM_2M_5}=\angle{BDC}$ and $\angle{BM_2M_6}=\angle{BDA}$ hence $$\angle{M_4M_2M_3}=\angle{BDA}+\angle{BDC}=\angle{ADC}=\angle{ABC}=\angle{M_4M_1M_3}$$

I complex bashed this problem

All angle are consider in mod 180 refer as Directed Angles :- \measuredangle. Let $P,Q,R,T,M,N$ be midpoint of $AC,BD,AE,CF,BC,BA$ respectively. From midpoint theorem $NQ \parallel AD$ and $MQ \parallel CD$ with $DA=DC \Rightarrow QM=QN$. As $\Box BNPM$ is parallelogram $\measuredangle{CDA}=\measuredangle{MQN}=\measuredangle{NBM}=\measuredangle{NPM}$ Hence $N,M,P,Q$ are cyclic. Claim: $\triangle RNQ \cong \triangle TMQ$ We will prove $\measuredangle{RNQ}=\measuredangle{TMQ}$ Note $\measuredangle{ANR}=\measuredangle{B},\measuredangle{ANQ}=\measuredangle{BNQ}=\measuredangle{BAD}$ hence $\Rightarrow $ $$\measuredangle{RNQ}=\measuredangle{ANQ}-\measuredangle{ANR}=\measuredangle{BAD}-\measuredangle{B}$$ and similarly $$\measuredangle{TMQ}=\measuredangle{QMC}-\measuredangle{TMC}=\measuredangle{B}-\measuredangle{DCB}$$ Now its well know property in $\triangle ABC$$\measuredangle{BAC}+\measuredangle{ACB}=\measuredangle{ABC}=\measuredangle{B}$\newline $$\measuredangle{TMQ}-\measuredangle{RNQ}=2\measuredangle{B}-(\measuredangle{BAC}+\measuredangle{CAD}+\measuredangle{ACB}+\measuredangle{DCA})=2\measuredangle{B}-(2\measuredangle{B})=0$$ hence we get $\measuredangle{RNQ}=\measuredangle{TMQ}$ from $AE=AF=2NR=2MT$ we prove our claim. Claim give us $\measuredangle{NQR}=\measuredangle{MQT} \Rightarrow \measuredangle{MQN}=\measuredangle{TQR}=\measuredangle{MPN}=\measuredangle{TPR}$ Hence $P,Q,R,T$ are cyclic. 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Om245 wrote: All angle are consider in mod 180 refer as Directed Angles :- \measuredangle. Let $P,Q,R,T,M,N$ be midpoint of $AC,BD,AE,CF,BC,BA$ respectively. From midpoint theorem $NQ \parallel AD$ and $MQ \parallel CD$ with $DA=DC \Rightarrow QM=QN$. As $\Box BNPM$ is parallelogram $\measuredangle{CDA}=\measuredangle{MQN}=\measuredangle{NBM}=\measuredangle{NPM}$ Hence $N,M,P,Q$ are cyclic. Claim: $\triangle RNQ \cong \triangle TMQ$ We will prove $\measuredangle{RNQ}=\measuredangle{TMQ}$ Note $\measuredangle{ANR}=\measuredangle{B},\measuredangle{ANQ}=\measuredangle{BNQ}=\measuredangle{BAD}$ hence $\Rightarrow $ $$\measuredangle{RNQ}=\measuredangle{ANQ}-\measuredangle{ANR}=\measuredangle{BAD}-\measuredangle{B}$$ and similarly $$\measuredangle{TMQ}=\measuredangle{QMC}-\measuredangle{TMC}=\measuredangle{B}-\measuredangle{DCB}$$ Now its well know property in $\triangle ABC$$\measuredangle{BAC}+\measuredangle{ACB}=\measuredangle{ABC}=\measuredangle{B}$\newline $$\measuredangle{TMQ}-\measuredangle{RNQ}=2\measuredangle{B}-(\measuredangle{BAC}+\measuredangle{CAD}+\measuredangle{ACB}+\measuredangle{DCA})=2\measuredangle{B}-(2\measuredangle{B})=0$$ hence we get $\measuredangle{RNQ}=\measuredangle{TMQ}$ from $AE=AF=2NR=2MT$ we prove our claim. Claim give us $\measuredangle{NQR}=\measuredangle{MQT} \Rightarrow \measuredangle{MQN}=\measuredangle{TQR}=\measuredangle{MPN}=\measuredangle{TPR}$ Hence $P,Q,R,T$ are cyclic. 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gliding away

Let $Q$ be the Miquel point of $AFEC$ and $D'$ be the reflection of $D$ across $AC$. Let $M$ and $N$ be such that $AMED$ and $CNFD$ are parallelograms. Let $EM$ and $FN$ intersect at $P$. Claim: $D'ABC$ concyclic $$\angle AD'C=\angle ADC=\angle ABC$$Claim: $P$ lies on $(BEF)$ $$\angle FPE=\angle ADC=\angle FBE$$Claim: $CAQD'$ is similar to $EFQB$ The spiral symmetry at $Q$ sends $EF$ to $AC$, since $BF=BE$ and $D'A=D'C$ we are done. Claim: $P$, $Q$, and $D'$ are collinear $$\angle QPE=\angle QBC=\angle QD'C=180^{\circ}-\angle EPD'$$Claim: Triangles $BQD'$, $BFN$, and $BEM$ are similar Notice $\angle BQD'=\angle BFN=\angle BEM$ and $$\frac{BQ}{QD'}=\frac{EF}{AC}=\frac{BF}{DC}=\frac{BF}{FN}=\frac{BE}{EM}$$ Thus $BD'MN$ is concyclic and taking a homothey centered at $D$ with ratio $1/2$ finishes the problem.