Problem 1673

1. **Reflecting $B$ over $AC$:**
Given that $\triangle ABC$ is a right-angled triangle with $\angle ABC = 90^\circ$, reflecting $B$ over $AC$ results in point $B'$. Since $\angle ABC = 90^\circ$, the reflection $B'$ will also form a right angle with $AC$. Therefore, $\angle AB'C = 90^\circ$.

2. **Midpoint $M$ and circle centered at $M$:**
Let $M$ be the midpoint of $AC$. Since $M$ is the midpoint, $M$ is equidistant from $A$ and $C$. The circle centered at $M$ with radius $MA = MC$ will pass through both $A$ and $C$.

3. **Point $B'$ on the circle:**
Since $\angle AB'C = 90^\circ$, $B'$ lies on the circle with diameter $AC$ (Thales' theorem). This is because any angle inscribed in a semicircle is a right angle.

4. **Choosing point $D$ on $\overrightarrow{BM}$:**
We choose point $D$ on $\overrightarrow{BM}$ such that $BD = AC$. Since $M$ is the midpoint of $AC$, $BM$ is perpendicular to $AC$ and $BD$ is equal to the length of $AC$.

5. **Point $D$ on the circle:**
Since $BD = AC$ and $B$ is on the circle with diameter $AC$, $D$ must also lie on this circle. This is because the length $BD$ is equal to the diameter of the circle, and thus $D$ must be on the circle.

6. **Congruence of triangles $\triangle BFC$ and $\triangle B'FC$:**
Let $F$ be the foot of the perpendicular from $B$ to $AC$. Since $B$ and $B'$ are reflections over $AC$, $BF = B'F$ and $\angle BFC = \angle B'FC = 90^\circ$. Therefore, $\triangle BFC \cong \triangle B'FC$ by the Hypotenuse-Leg (HL) congruence theorem.

7. Angle bisector property:
Since $\triangle BFC \cong \triangle B'FC$, we have $\angle BCF = \angle B'CF$. This implies that $\angle B'CF$ is bisected by $CF$. Since $F$ is the midpoint of $AC$, $CF$ is the angle bisector of $\angle B'CD$.

8. Inscribed angles:
Since $B'$ and $D$ lie on the circle with diameter $AC$, the inscribed angles $\angle B'CA$ and $\angle DCA$ are equal. Therefore, $\angle B'CA = \angle DCA$.

9. Conclusion:
Since $\angle B'CA = \angle DCA$, $B'C$ is the angle bisector of $\angle MB'D$.

$\blacksquare$