What is the value of segment AB in the quadrilateral of the question below? 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)\\
$
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Using your notation,

*

*Angle Bisector Theorem on $\triangle AHD$ yields $$\overline{AH} = \frac{12}7 \overline{HC}.$$

*Note that $BC$ bisects $\angle HCA$. Angle Bisector Theorem on $\triangle HCA$ gives, together with 1. $$\overline{AC} = \frac{\overline{AB}\cdot \overline{HC}}{\frac{12}7\overline{HC}-\overline{AB}}.$$

*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.

