The parallel lines $\overset{\longleftrightarrow}{AB} \parallel \overset{\longleftrightarrow}{CD}$ are cut by a transversal $\overset{\longleftrightarrow}{EF}.$

The angles $\angle{JID}$ and $\angle{IJA}$ are interior to the parallel lines and are on opposite sides of the transversal. So, they are alternate interior angles, and consequently, they have the same measure.

Setting the measures of $\angle{JID}$ and $\angle{IJA}$ equal to each other, we have an equation that we can solve for $x\mathbin{:}$

\begin{align*} \require{cancel} m\angle{JID} &= m\angle{IJA} \\[5pt] 108^\circ &= 3x +24^\circ \\[5pt] 84^\circ &= 3x \\[5pt] \dfrac{84^\circ}{3} &= \dfrac{3x}{3} \\[5pt] 28^\circ &= \dfrac{\cancel{3}x}{\cancel{3}} \\[5pt] 28^\circ &= x \\[5pt] x &= 28^\circ \end{align*}

The parallel lines $\overset{\longleftrightarrow}{AB} \parallel \overset{\longleftrightarrow}{CD}$ are cut by a transversal $\overset{\longleftrightarrow}{EF}.$

The angles $\angle{JID}$ and $\angle{IJA}$ are interior to the parallel lines and are on opposite sides of the transversal. So, they are alternate interior angles, and consequently, they have the same measure.

Setting the measures of $\angle{JID}$ and $\angle{IJA}$ equal to each other, we have an equation that we can solve for $x\mathbin{:}$

\begin{align*} \require{cancel} m\angle{JID} &= m\angle{IJA} \\[5pt] 2x + 55^\circ &= 4x - 25^\circ \\[5pt] 55^\circ &= 2x - 25^\circ \\[5pt] 80^\circ &= 2x \\[5pt] \dfrac{80^\circ}{2} &= \dfrac{2x}{2} \\[5pt] 40^\circ &= \dfrac{\cancel{2}x}{\cancel{2}} \\[5pt] 40^\circ &= x \\[5pt] x &= 40^\circ \end{align*}

The parallel lines $\overset{\longleftrightarrow}{AB} \parallel \overset{\longleftrightarrow}{CD}$ are cut by a transversal $\overset{\longleftrightarrow}{EF}.$

The angles $\angle{DIF}$ and $\angle{AJE}$ are exterior to the parallel lines and are on opposite sides of the transversal. So, they are alternate exterior angles, and consequently, they have the same measure.

Setting the measures of $\angle{DIF}$ and $\angle{AJE}$ equal to each other, we have an equation that we can solve for $x\mathbin{:}$

\begin{align*} \require{cancel} m\angle{DIF} &= m\angle{AJE} \\[5pt] 120^\circ &= 6x+12^\circ \\[5pt] 108^\circ &= 6x \\[5pt] 6x &= 108^\circ \\[5pt] \dfrac{6x}{6} &= \dfrac{108^\circ}{6} \\[5pt] \dfrac{\cancel{6}x}{\cancel{6}} &= 18^\circ \\[5pt] x &= 18^\circ \end{align*}

The parallel lines $\overset{\longleftrightarrow}{AB} \parallel \overset{\longleftrightarrow}{CD}$ are cut by a transversal $\overset{\longleftrightarrow}{EF}.$

The angles $\angle{FIC}$ and $\angle{EJB}$ are exterior to the parallel lines and are on opposite sides of the transversal. So, they are alternate exterior angles, and consequently, they have the same measure.

Setting the measures of $\angle{FIC}$ and $\angle{EJB}$ equal to each other, we have an equation that we can solve for $x\mathbin{:}$

\begin{align*} \require{cancel} m\angle{FIC} &= m\angle{EJB} \\[5pt] 5x-12^\circ &= 180^\circ -x \\[5pt] 6x-12^\circ &= 180^\circ \\[5pt] 6x &= 192^\circ \\[5pt] \dfrac{6x}{6} &= \dfrac{192^\circ}{6} \\[5pt] \dfrac{\cancel{6}x}{\cancel{6}} &= 32^\circ \\[5pt] x &= 32^\circ \end{align*}

Consider $\angle 2$ in the diagram below.

Notice that $\angle 2$ and the angle whose measure is $55^\circ$ are alternate exterior angles formed when the transversal $u$ cuts the parallel lines $s$ and $t.$ Hence,

\[ m\angle 2 = 55^{\circ}\,. \]

Also, notice that the angle determined by $\angle 1$ and $\angle 2$ and the angle whose measure is $127^\circ$ are corresponding angles formed when the transversal $v$ cuts the parallel lines $r$ and $s.$ Hence,

\[ m\angle 1 + m\angle 2 = 127^{\circ}\,. \]

Now, substituting $m\angle 2$ into the equation, we can find $m\angle1 \mathbin{:}$

\begin{align*} m\angle 1 + m\angle 2 &= 127^{\circ} \\[3pt] m\angle 1 + 55^\circ &= 127^{\circ} \\[3pt] m\angle 1 &= 72^{\circ} \end{align*}

Consider $\angle 2$ in the diagram below.

Notice that $\angle 2$ and the angle whose measure is $72^\circ$ are alternate exterior angles formed when the transversal $v$ cuts the parallel lines $r$ and $s.$ Hence, \[ m\angle 2 = 72^{\circ}. \]

Also, notice that the angle determined by $\angle 1$ and $\angle 2$ and the angle whose measure is $139^\circ$ are corresponding angles formed when the transversal $u$ cuts the parallel lines $s$ and $t.$ Hence, \[ m\angle 1 + m\angle 2 = 139^{\circ}. \]

Now, substituting $m\angle 2$ into the equation, we can find $m\angle1 \mathbin{:}$ \begin{align*} m\angle 1 + m\angle 2 & = 139^{\circ} \\[5pt] m\angle 1 + 72^\circ & = 139^{\circ} \\[5pt] m\angle 1 & = 67^{\circ} \end{align*}

Notice that $\angle{UGV}$ and $\angle{GVH}$ are alternate interior angles formed when the transversal $\overleftrightarrow{VG}$ cuts the parallel lines $\overleftrightarrow{HQ}$ and $\overleftrightarrow{GC}.$ Hence,

\[ m\angle{GVH} = m\angle{UGV} = 46^{\circ}\,. \]

Also, notice that $\angle{VUC}$ and $\angle{HVU}$ are alternate interior angles formed when the transversal $\overleftrightarrow{UV}$ cuts the parallel lines $\overleftrightarrow{HQ}$ and $\overleftrightarrow{GC}.$ Hence,

\[ m\angle{HVU} =m\angle{VUC} = 118^{\circ}\,. \]

Now, using the angle addition postulate, we can find $\angle{GVU}\mathbin{:}$

\begin{align*} m \angle{GVU} &=m \angle{HVU} - m \angle{GVH} \\[3pt] & = 118^{\circ}-46^{\circ} \\[3pt] &= 72^{\circ} \end{align*}

Notice that $\angle{BAF}$ and $\angle{AFC}$ are alternate interior angles formed when the transversal $\overleftrightarrow{AF}$ cuts the parallel lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{CF}.$ Hence, \[ m\angle{AFC} = m\angle{BAF} = 71^{\circ}. \]

Also, notice that $\angle{FBE}$ and $\angle{BFC}$ are alternate interior angles formed when the transversal $\overleftrightarrow{BF}$ cuts the parallel lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{CF}.$ Hence, \[ m\angle{BFC} =m\angle{FBE} = 139^{\circ}. \]

Now, using the angle addition postulate, we can find $\angle{AFB}{:}$ \begin{align*} m \angle{AFB} & = m \angle{BFC} - m \angle{AFC} \\[5pt] & = 139^{\circ} - 71^{\circ} \\[5pt] & = 68^{\circ} \end{align*}