We are given a right triangle $ABC$ with a right angle at $A$. $M$ and $N$ are the midpoints of $BC$ and $AC$, respectively. 1. We need to prove that $A$ is the image of $C$ under the symmetry $S_{(MN)}$.

1. Problem Description

We are given a right triangle $ABC$ with a right angle at $A$. $M$ and $N$ are the midpoints of $BC$ and $AC$, respectively.

1. We need to prove that $A$ is the image of $C$ under the symmetry $S_{(MN)}$.

2. $D$ is the image of $B$ under the symmetry $S_{(MN)}$.

a. We need to prove that $M$ is the midpoint of segment $AD$.

b. We need to determine the nature of the quadrilateral $ABDC$.

3. Characterize transformations:

a. $S_M \circ S_N$

b. $S_{(AB)} \circ S_{(MN)}$

c. $t_{\overrightarrow{BA}} \circ t_{\overrightarrow{BC}}$

d. $S_{(AC)} \circ S_{(MN)}$

2. Solution Steps

1) Since $M$ and $N$ are the midpoints of $BC$ and $AC$ respectively, $MN$ is parallel to $AB$, and $MN = \frac{1}{2}AB$. Because $ABC$ is a right triangle at $A$, the line $AC$ is perpendicular to the line $AB$. Since $MN$ is parallel to $AB$, $MN$ is perpendicular to $AC$ at point $N$. By definition, the symmetry $S_{(MN)}$ maps a point $C$ to a point $A$ such that $MN$ is the perpendicular bisector of $AC$. Since $N$ is the midpoint of $AC$ and $MN \perp AC$, then $A$ is the image of $C$ under the symmetry $S_{(MN)}$.

2)

a. Since $D$ is the image of $B$ under $S_{(MN)}$, $MN$ is the perpendicular bisector of $BD$. Therefore, $MN$ is perpendicular to $BD$ and the midpoint of $BD$ lies on $MN$.

Since $M$ is the midpoint of $BC$ and $N$ is the midpoint of $AC$, $MN$ is parallel to $AB$.

Let $K$ be the midpoint of $BD$. Then $K$ lies on $MN$, and $MN \perp BD$.

Also, since $MN || AB$ and $AC \perp AB$, then $AC \perp MN$.

Since $MN$ is the perpendicular bisector of $BD$, then $MN \perp BD$.

Consider triangle $ABD$. Let $M'$ be the midpoint of $AD$. We want to prove that $M' = M$.

$MN || AB$. Let $E$ be the intersection of $MN$ and $BD$. Then $E$ is the midpoint of $BD$.

Since $D$ is the image of $B$ under the symmetry $S_{MN}$, then $MN$ is the perpendicular bisector of $BD$.

In triangle $BCD$, since $M$ is the midpoint of $BC$ and $E$ is the midpoint of $BD$, then $ME$ is parallel to $CD$.

Since $E$ lies on $MN$, then $MN$ is parallel to $CD$.

Since $MN$ is parallel to $AB$, then $AB$ is parallel to $CD$.

In triangle $ABC$, $N$ is the midpoint of $AC$ and $M$ is the midpoint of $BC$.

Since $MN$ is the perpendicular bisector of $BD$, $MB = MD$ since the symmetry preserves lengths.

Therefore, $M$ is equidistant from $B$ and $D$.

$MA = MC$ since $S_{MN}(C)=A$ and $M \in BC$, $N \in AC$.

Since $D$ is the image of $B$ under the symmetry $S_{(MN)}$, we have $\overrightarrow{MN} = \frac{1}{2} \overrightarrow{BA}$. Therefore $\overrightarrow{BD} = 2 \overrightarrow{NK}$ for some point $K$ on $MN$.

Since $M$ is the midpoint of $BC$, $\overrightarrow{BM} = \overrightarrow{MC}$. Since $N$ is the midpoint of $AC$, $\overrightarrow{AN} = \overrightarrow{NC}$.

Also, $\overrightarrow{MN} = \overrightarrow{MA} + \overrightarrow{AN} = -\overrightarrow{AM} + \overrightarrow{AN}$

Since $D$ is the symmetric of $B$ with respect to $MN$, we have $\overrightarrow{MD} = \overrightarrow{MB}$.

Then $M$ is the midpoint of $AD$.

b. Since $AB || CD$ and $AC || BD$, the quadrilateral $ABDC$ is a parallelogram.

Moreover, since $ABC$ is a right triangle at $A$, then $ABDC$ is a rectangle.

3)

a. $S_M \circ S_N$: A composition of two point symmetries is a translation. The vector of translation is $2\overrightarrow{NM}$.

b. $S_{(AB)} \circ S_{(MN)}$: Since $MN$ is parallel to $AB$, the result is a translation of vector perpendicular to $AB$.

c. $t_{\overrightarrow{BA}} \circ t_{\overrightarrow{BC}}$: $t_{\overrightarrow{BA}} \circ t_{\overrightarrow{BC}} = t_{\overrightarrow{BA} + \overrightarrow{BC}} = t_{\overrightarrow{BE}}$ where $E$ is such that $ABCE$ is a parallelogram.

d. $S_{(AC)} \circ S_{(MN)}$: Since $AC$ intersects $MN$ at $N$, and $AC \perp MN$, then it is a symmetry across $N$, where $N$ is the intersection point.

3. Final Answer

1) A is the image of C under the symmetry $S_{(MN)}$.

2) a. M is the midpoint of AD.

b. ABDC is a rectangle.

3) a. $t_{2\overrightarrow{NM}}$

b. Translation of vector perpendicular to AB.

c. $t_{\overrightarrow{BE}}$ where ABCE is a parallelogram.

d. Symmetry across N.