What is the value of segment AB in the quadrilateral of the question below?

Question

For reference: (exact copy of question): In the $ABCD$ quadrilateral, the diagonals intersect at $Q$. In the prolongation of $BC$, the point F is marked: $\angle BCA=\angle FCD, \angle BAC=\angle CAD, AD=12, QC=3, CD=7$. Calculate AB. (answer: $9$)

My progress: Follow the figure according to the statement. I didn't get great evolutions...I didn't get great evolutions...maybe there is some theorem that can help angle bisector theorem: $\triangle ABD:\frac{AB}{BQ} =\frac{AD}{DQ}\\ \triangle AHD: \frac{AH}{HC}=\frac{12}{7}$

T.Menelao: $AB.HC.DG = BH.CD.AG\\ AB.HC.DG = BH.7.(12+DG)\\ $ ...

Answer 1

Using your notation,

1. Angle Bisector Theorem on $\triangle AHD$ yields $$\overline{AH} = \frac{12}7 \overline{HC}.$$
2. Note that $BC$ bisects $\angle HCA$. Angle Bisector Theorem on $\triangle HCA$ gives, together with 1. $$\overline{AC} = \frac{\overline{AB}\cdot \overline{HC}}{\frac{12}7\overline{HC}-\overline{AB}}.$$
3. Menelaus's Theorem on $\triangle HCA$ with secant $BD$ yields $$(\overline{AC}-3) \cdot 7 \cdot (\overline{AH}-\overline{AB}) = 3\cdot (\overline{HC} + 7) \cdot \overline{AB}.$$Use now 1. and 2. in the previous equation to get $$7\left(\frac{\overline{AB}\cdot \overline{HC}}{\frac{12}7\overline{HC}-\overline{AB}}-3\right)\left(\frac{12}7\overline{HC}-\overline{AB}\right)=3\cdot (\overline{HC} + 7) \cdot \overline{AB}$$which is equivalent to $$\require{cancel} 7\overline{AB}\cdot \overline{HC}-36\overline{HC}+\cancel{21\overline{AB}}=3\overline{AB}\cdot \overline{HC}+\cancel{21\overline{AB}}. $$ Simplifying and dividing by $\overline{HC}$ (which is not $0$) leads to the result.

Answer 2

The question is saying that F is at any point on the prolongation of BC. So I took F as the intersection of BC and the line parallel to AC.

Let's say $\angle BCA = \angle DCF = a$ and $\angle BAC = \angle CAD = b$. Then DF = a. You can now get the $\frac{BQ}{BD}=\frac{3}{7}$.

Now, let's say $BQ=3x$ and $QD = 4x$ to maintain the above ratio.

As AQ is the angle bisector of $\angle BAD$ in $\triangle BAD$

$\frac{AB}{12}=\frac{3}{4}$

By simplifying you get,

$\boxed{AB=9}$