Since two sides are congruent, $\overline{AC} \cong \overline{BC},$ we conclude that $\triangle ABC$ is isosceles. Consequently, the base angles must have the same measure:

\[ m\angle B = m\angle A = 2x+28^\circ \]

Now, since the internal angles of any triangle sum to $180^\circ,$ we have \begin{align*} m\angle A + m\angle B + m\angle C &= 180^{\circ}\\[2pt] \left( 2x+28^\circ\right) + \left( 2x+28^\circ\right) + \left( 3x-23^{\circ}\right) &= 180^{\circ}\\[2pt] 7x+33^{\circ} &=180^{\circ}\\[2pt] 7x &=147^{\circ}\\[2pt] x &= 21^{\circ}\,. \end{align*}

Finally, substituting $x=21^\circ$ into the equation for the measure of $\angle B$ gives \begin{align*} m\angle{B} &= m\angle{A}\\[2pt] &= 2x+28^{\circ}\\[2pt] &= 2(21^{\circ}) +28^{\circ}\\[2pt] &= 70^{\circ}\,. \end{align*}

So $m\angle A = 70^\circ.$

Since two sides are congruent, $\overline{BC} \cong \overline{AC},$ we conclude that $\triangle ABC$ is isosceles. Consequently, the base angles must have the same measure:

\[ m\angle B= m\angle A = x +40^\circ \]

Now, since the internal angles of any triangle sum to $180^\circ,$ we have \begin{align*} m\angle{A} + m\angle{B} + m\angle{C} &= 180^\circ \\[2pt] (x+40^{\circ}) + (x+40^{\circ}) + (2x-12^{\circ}) &= 180^{\circ}\\[2pt] 4x &=112^{\circ}\\ x &= 28^{\circ}\,. \end{align*}

Finally, substituting $x=28^\circ$ into the equation for the measure of $\angle C$ gives \begin{align*} m\angle{C} &= 2x-12^{\circ} \\[2pt] &= 2(28^{\circ}) -12^\circ \\[2pt] &= 44^{\circ}\,. \end{align*}

So $m\angle C = 44^\circ.$

First, from the diagram, notice that $\angle{BCA}$ and $\angle{BCD}$ are supplementary. Therefore, \begin{align*} m\angle{BCA} + m\angle{BCD} &= 180^\circ \\[3pt] m\angle{BCA}+ y &= 180^\circ \\[3pt] m\angle{BCA} &= 180^\circ - y. \end{align*}

Since $\triangle{ABC}$ is an isosceles triangle where $\overline{AC} \cong \overline{BC}$, the base angles must be equal:

\[ m\angle{B} = m\angle{A}=69^\circ. \]

Finally, since the interior angles of any triangle sum to $180^\circ,$ we have \begin{align*} m\angle{A} + m\angle{B} + m\angle{BCA} &= 180^\circ \\[3pt] 69^\circ + 69^\circ + (180^\circ - y) &= 180^\circ \\[3pt] y &= 138^\circ. \end{align*}

First, from the diagram, notice that $\angle{CBA}$ and $\angle{DBC}$ are supplementary. Therefore, \begin{align*} m\angle{CBA} + m\angle{DBC} &= 180^\circ \\[3pt] m\angle{CBA} + 110^\circ &= 180^\circ \\[3pt] m\angle{CBA} &= 70^\circ . \end{align*}

Since $\triangle{ABC}$ is an isosceles triangle, where $\overline{BC} \cong \overline{CA}$, the base angles must be equal: \[ m\angle{A} = m\angle{CBA}=70^\circ. \]

Finally, since the interior angles of any triangle sum to $180^\circ,$ we have

\begin{align*} m\angle{A} + m\angle{CBA} + m\angle{C} &= 180^\circ \\[3pt] 70^\circ + 70^\circ + m\angle{C} &= 180^\circ \\[3pt] 140^\circ +m\angle{C} &= 180^\circ \\[3pt] m\angle{C} &= 40^\circ. \end{align*}

Since $\triangle ABC$ is an isosceles triangle with base $\overline{BC}$, we have that $\overline{AB} \cong \overline{AC}.$ Hence, \begin{align*} {AB}&={AC}\\[3pt] 4x-7 &= 2x+5\\[3pt] 2x &= 12\\[3pt] x &= 6\,. \end{align*}

Substituting this value into the expression for the measure of $\overline{BC}$ gives \begin{align*} {BC} &= 3x-3 \\[3pt] &= 3(6) -3 \\[3pt] &= 15\,. \end{align*}

Since $\triangle{ABC}$ is an isosceles triangle with base $\overline{AB}$, we have that $\overline{AC} \cong \overline{CB}.$ Hence, \begin{align*} AC &= CB\\[3pt] 2x-1 &=x+4\\[3pt] x &= 5\,. \end{align*}

Substituting this value into the expression for the measure of $\overline{AC}$ gives \begin{align*} AC &= 2x-1 \\[3pt] &= 2(5) -1 \\[3pt] &= 9\,. \end{align*}

First, notice that $\triangle{ABC}$ is an isosceles triangle. Consequently, the base angles must have equal measure:

\[ m\angle{A}=m\angle{ABC}=68^\circ \]

Also, since $\triangle{CDB}$ is an isosceles triangle, we have:

\[ m\angle{ D}=m\angle{BCD} \]

Since $ \angle {CBA}$ and $ \angle {CBD}$ are supplementary, we have \begin{align*} m \angle {ABC} +m \angle {CBD}&= 180^\circ\\[3pt] 68^\circ+m \angle {CBD} &= 180^\circ\\[3pt] m \angle {CBD} &=112^\circ. \end{align*}

Now, we calculate $m\angle {D}$ using the fact that the interior angles of a triangle sum to $180^\circ\mathbin{:}$ \begin{align*} m \angle {D} + m \angle {BCD} + m \angle {CBD} &= 180^\circ \\[3pt] m \angle {D} + m \angle {D} + 112^\circ &= 180^\circ \\[3pt] 2m \angle {D}&= 68^\circ \\[3pt] m \angle {D} &= 34^\circ \end{align*}

First, we calculate $m\angle {D}$ using the fact that the interior angles of $\triangle{DBC}$ sum to $180^\circ\mathbin{:}$ \begin{align*} m\angle{D} + m\angle{DBC} + m\angle{BCD} &= 180^\circ \\[3pt] m\angle{D} + 75^\circ + 30^\circ &= 180^\circ\\[3pt] m\angle D + 105^\circ&= 180^\circ\\[3pt] m\angle D &= 75^\circ \end{align*}

Since $m\angle D=m\angle CBD=75^\circ$, we have that $\angle D\cong\angle CBD$ and therefore \[ \overline{CB} \cong \overline{CD}. \]

Thus $\triangle{DBC}$ is isosceles.

Since $\overline{AB} \cong \overline{CD}$ and $\overline{CD} \cong \overline{BC}$, then \[ \overline{AB} \cong \overline{BC}. \] Hence, $\triangle{ABC}$ is also isosceles. Therefore, we have \[ m\angle ACB = m\angle A \]

Finally, we calculate $m\angle {A}$ using the fact that the interior angles of $\triangle{ABC}$ sum to $180^\circ\mathbin{:}$ \begin{align*} m\angle{A} + m\angle{CBA} + m\angle{ACB} &= 180^\circ \\[3pt] m\angle A + 70^\circ + m\angle A &= 180^\circ\\[3pt] 70^\circ + 2 m\angle A &= 180^\circ\\[3pt] 2 m\angle A &= 110^\circ\\[3pt] m\angle A &= 55^\circ. \end{align*}