lema 1: $ED \parallel E'D'$ proof: $EH_1=E'H_1$ $DH_2=D'H_2$ $\implies$ $EH_1=E'H_1=DH_2=D'H_2$ $\implies$ $ED \parallel E'D'$ $H_{1}H_{2} \parallel ED$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \blacksquare$ lema2:$SX=XD'$ proof: $XH_2 \parallel SD $ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\stackrel{ thales}{\implies}$ $SX=XD'$ $D'H_2=H_2 D $ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \blacksquare$ we create a new E' such this: reflection $T$ over $x$ and named $E''$ we must prove that $E''=E'$ lema3:$ED \parallel E''D'$ proof: $\measuredangle E''XD=\measuredangle SXT$ $XD'=XS \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \implies$ $XST \cong E''D'X \implies ED \parallel E''D'$ $XE''=TX$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \blacksquare$ so $E''$ lie on $E'D'$ line we have $ED=E''D' , ST=E''D'$ so we must prove that $ED=ST$ and for prove this: $\frac{ES}{AX}=\frac{BE}{BA}=\frac{CD}{CA}=\frac{DT}{AX}$ so we have $\frac{ES}{AX}=\frac{DT}{AX}$ $\implies ES=DT \implies ES+SD=DT+SD=ST=ED$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \blacksquare$ so $E''=E'$

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solved also here

It has a easy but nice solution via Trig-Ceva.

Let $F$ be the intersection of $E'B , D'C$ It is clear that $D'E' =// DE$ , since $D'E'$ is the reflection of $DE$ over the line $l$ Let $h_{1} , h , h'_{1} , h'$ be the heights of triangles $ADE , ABC , FD'E' , FBC$ from vertices $A , A , F , F$ respectively. Then , since $\triangle ADE \sim ABC$ and $FD'E' \sim FBC \Rightarrow \frac{h_{1}}{h} = \frac{h'_{1}}{h'} = \frac{DE}{BC} = \frac{DE'}{BC}$ But since $h + h_{1} = h' + h'_{1} \Rightarrow h_{1} = h'_{1} \Rightarrow AF // BC \Rightarrow AF \equiv l$ , as desired

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JustARandomGuy__ wrote: Let $F$ be the intersection of $E'B , D'C$ It is clear that $D'E' =// DE$ , since $D'E'$ is the reflection of $DE$ over the line $l$ Let $h_{1} , h , h'_{1} , h'$ be the heights of triangles $ADE , ABC , FD'E' , FBC$ from vertices $A , A , F , F$ respectively. Then , since $\triangle ADE \sim ABC$ and $FD'E' \sim FBC \Rightarrow \frac{h_{1}}{h} = \frac{h'_{1}}{h'} = \frac{DE}{BC} = \frac{DE'}{BC}$ But since $h + h_{1} = h' + h'_{1} \Rightarrow h_{1} = h'_{1} \Rightarrow AF // BC \Rightarrow AF \equiv l$ , as desired splendid solution!

Generalized problem: Let $D$ and $E$ on $AC$ and $AB$ respectively , a line $l$ through $A$ such that $l ,BC,DE$ share a point ( finite point or infinity point); let $D'$ and $E'$ be the reflections of $D$ and $E$ wrt $l$ respectively . Prove that $D'B, E'C$ and $l$ are concurrent. The proof based on harmonic bundles.