Construction of symmedians but different. Given $2$ lines $l_1$ and $l_2$ intersecting at point $F,$ let $A$ and $B$ on line $l_1,$ construct point $E$ on $l_2$ such that $\angle AEF=\angle (BE,EF) .$
Here's the diagram of  the problem:

For the construction, I thought of taking point $D$ such that $BD\perp DF.$ But couldn't proceed much.
 A: 
Let $P$ and $Q$ be points on the line $l_{2}$ such that $AP\perp l_{2} $ and $BQ\perp l_{2}$. Extend $AP$ and $BQ$ to points $R$ and $S$ such that $AP=PR$ and $BQ=QS$. Now draw $AS$ and $BR$. These two segments will intersect at some point on the line $l_{2}$.That point is our desired point $E$.

Reason : Let $BR$ intersect $l_{2}$ in point $T$.Draw $AT$ and $ST$. Observe that $\triangle ATR$ and $\triangle BTS$ are both isosceles and $\angle ART=\angle SBT$; Thus, $\triangle ATR\sim \triangle BTS$. Hence $\angle ATP=\angle STQ$ and thereafter $AS$ will intersect $l_{2}$ in point $T$.Now $\angle ATP=\angle BTQ$ and hence Point $T$$=$ Point $E$.
A: Let $C$ be symmetric to $B$ with respect to $l_2$.
Then the line $AC$ intersects $l_2$ at $E$ you seek.
Proof of correctness:
Let $K$ be the intersection of $l_2$ and $BC$. Then triangles $\triangle EKB$ and $\triangle EKC$ are congruent by the SAS rule (angles at $K$ are both right, hence equal, and $BK=KC$ by symmetry). Consequenlty their angles at $E$ are equal. Finally, the angle $\angle AEF$ is opposite to $\angle CEK$, so congruent to it and to $\angle KEB$, as required.
A: The construction:
Take any point $D$ on $l_2$. Now intersect $AD$ and $BD$ with any line through $F$ to get two points $X$ and $Y$ respectively. Now intersect $AY$ and $BX$ at $P$. Then intersect $DP$ with $l_1$ at $Z$. Draw the circle with diameter $FZ$. Then the intersection of this circle with $l_2$ is our point $E$.
Proof that it works:
The point $Z$ is just the harmonic conjugate of $F$ with respect to segment $AB$, so the circle with diameter $FZ$ is the Apollonian circle through $F$. It's well known that for any point $E'$ on this circle, $\angle AE'B$ is bisected by $E'Z$, and more importantly $E'F$ is the external angle bisector of $\angle AE'B$. So intersecting this with $l_2$ gives the desired point.

