1. Construct Points and Define Angles:
- Let $AL$ be the internal bisector of $\triangle ABC$ with $L \in BC$.
- Choose point $K$ on segment $AL$.
- Point $P$ lies on the same side with respect to line $BC$ as point $A$ such that $\angle BPL = \angle CKL$ and $\angle CPL = \angle BKL$.
- Let $M$ be the midpoint of segment $KP$.
- Let $D$ be the foot of the perpendicular from $K$ to $BC$.

2. Construct Similar Triangles:
- Construct points $E$ and $F$ such that $\triangle EBL \sim \triangle KLC$ and $\triangle FCL \sim \triangle KLB$.
- Note that $P$ is the second intersection of $(EBL)$ and $(FCL)$.
- Let $Q = (EBLP) \cap \overline{AB}$ and $R = (FCLP) \cap \overline{AC}$.
- Let $N$ be the reflection of $L$ over $D$.

3. Apply Miquel Theorem:
- Observe that $P$ lies on $(AQR)$ by Miquel theorem.

4. Prove Collinearity of Points:
- Claim: $E, K, P, F$ are collinear.
- Proof:
- Collinearity of $P, E$, and $F$ comes from the angle chase:
$$\measuredangle EPL = \measuredangle EBL = \measuredangle KLC = \measuredangle FCL = \measuredangle FPL$$
- Collinearity of $K, E$, and $F$ comes from the similarity:
$$\frac{KL}{EB} = \frac{LC}{BL} = \frac{FC}{KL}$$
This implies collinearity along with $KL \parallel BE \parallel CF$ (a well-known lemma).

5. Prove Similarity of Triangles:
- Claim: $\triangle ARK \sim \triangle ALB$.
- Proof:
- $K$ lies on $(AQRP)$ as $\measuredangle KPR = \measuredangle FPR = \measuredangle CFR = \measuredangle KAR$.
- Since $\measuredangle ARL = \measuredangle CRL = \measuredangle CPL = \measuredangle LKB = \measuredangle AKB$ and $\measuredangle BAK = \measuredangle LAR$, we conclude that $\triangle ABK \sim \triangle ALR$.
- Therefore, $A$ is the center of spiral similarity sending $\overline{BK}$ to $\overline{LR}$, and hence $A$ is also the center of spiral similarity sending $\overline{BL}$ to $\overline{KR}$.

6. Angle Bisector and Similarity:
- $\measuredangle APK = \measuredangle ARK = \measuredangle ALB = \measuredangle FCL = \measuredangle FPL = \measuredangle KPL \implies \overline{KP}$ bisects $\angle APL$.
- Thus,
$$\frac{AP}{PL} = \frac{AK}{KL} = \frac{AK}{KN}$$
- Additionally, $\measuredangle AKN = \measuredangle LKN = 2\measuredangle ALB = 2\measuredangle APK = \measuredangle APL$, hence $\triangle AKN \sim \triangle APL$.

7. Final Similarity and Angle Calculation:
- By the gliding principle, as $M, D$ are the midpoints of $\overline{KP}, \overline{NL}$ respectively, $\triangle AKN \sim \triangle APL \sim \triangle AMD$.
- Finally,
$$\measuredangle AMD = \measuredangle APL = 2\measuredangle ALB = \measuredangle ACB + \measuredangle ABC$$

Therefore, we have proven that $\angle AMD = 180^\circ - |\angle ABC - \angle ACB|$.

$\blacksquare$