Question:
As shown in the diagram, in isosceles $\triangle ABC,\ AB=AC.$ $H$ is a point on $AC$, take points $E$ and $F$ in turn on the extension line of $BC$, and take point $BD$ on the extension line of $CB$, so that $EF=DB$, make $EG // AC$ across point $E$ to cross the extension line of $DH$ at point $G$, connect $AF$, and if $\angle HDF+ \angle F= \angle BAC$: Explore the quantitative relationship between $\angle BAF$ and $\angle CHG$, and find a line in the figure that is equal to the line segment $AF$ and justify your conclusion.
(Note: The equivalent line must be only in the original image and cannot equivalent auxiliary lines.)
Attachment A: Image of the question
This question will be simple by using congruent triangles. But I still wonder how to deal with this question.
One possible way I've got is to:
For the "quantitative relationship between $\angle BAF$ and $\angle CHG$" part, I think $\angle BAF$ = $\angle CHG$, because $∠F + ∠HDF = ∠BAC$, it follows that $∠CHG = ∠FDG + ∠DCH = ∠HDF + ∠F + ∠CAF = ∠BAC + ∠CAF = ∠BAF. $
Detailed proving process:
$\because ∠HDF+∠F=∠BAC,$ $\therefore ∠CHG= ∠FDG+ ∠DCH= ∠HDF+∠F+ ∠CAF=∠BAC+∠CAF=∠BAF,$ $\therefore ∠CHG=∠BAF.$
For the "equivalent line of $AF$" part, I guess the equivalent line of $AF$ is $DE$. Extend $BD$ to $R$ such that $BR = CF$, connect $AR$, and make $AJ//CF$ to intersect the prolongation line of $EG$ at $J$, connecting $FJ$. Prove that the quadrilateral $ACEJ$ and quadrilateral $AJDR$ are parallelograms (for the latter, I partially know the proving process, so please write the detailed proving process for the conclusion "$AJDR$ is a parallelogram"). After this step, I didn't find ways to continue proving.
However, it is clear that it is very hard to work out because this is a classical "bonus/grand finale" question (压轴题) of the Chinese secondary school entrance examination ("zhongkao"/中考). Additionally, it is a congruent triangle "side, side, angle" ($SSA$) model (边边角模型) that is non-existent in congruent figure proving methods (they are $SSS, SAS, ASA, AAS,$ and $HL$).
This is a classical "bonus question" (or grand finale question, yazhou/yāzhóu question), which is known as "压轴题" (or "壓軸題") in Chinese, the name of the very very difficult examination question in China which "can open up students' mark and performance".
Here are the definitions of the term "bonus/grand finale question" in a Chinese encyclopedia:
"Generally refers to the big question that appears at the end of the examination paper. There are bonus questions in the formal examinations of mathematics and physics. These questions are generally of high marks and difficulty, testing strong general ability, and can open up the student's performance and mark in the examination, and are also the key study items for many students and teachers. "
I tried to work it out with better and easier methods but failed. Are there any possible approaches? Are research efforts provided by me right? Thank you for your ideas in advance!
Attachment B: Auxiliary lines I added in my research effort
$DJ=RA=AF$ (you already have this): Draw the perpendicular from $AX$ onto $BC$, so that $X$ lies on $BC$. Then, by AAS congruency, you have that $X$ is the midpoint of $BC$. So that $XB = XC$, and since $BR = CF$, you have $RX = BR + BX = CF + CX = FX$. Then by $SAS$ congruency, $RA = AF$. Since $RAJD$ is a parallelogram, its opposite sides are equal. So, $DJ = RA = AF$.
Consider $\triangle DJG$. $$\angle DJG=\angle RAC = \angle BAF = \angle CHG = \angle DGJ$$ where the first and last equalities follow because of parallel lines, and you have already proven the third. For the second, by the two congruencies mentioned above, you have $\angle RAX = \angle FAX$ and $\angle CAX = \angle BAX$. Add them to get $\angle RAC = BAF$.
Then, since $\angle DGJ = \angle DJG$, $DG = DJ$.