Given a triangle $ABC$ which is right-angled at $A$. $M$ and $N$ are the midpoints of segments $BC$ and $AC$ respectively. 1) Show that $A$ is the image of $C$ under the symmetry $S_{(MN)}$. 2) Point $D$ is the image of point $B$ under the symmetry $S_{(MN)}$. a) Show that $M$ is the midpoint of segment $AD$. b) What is the nature of quadrilateral $ABDC$? 3) Characterize: a) $S_M \circ S_N$ b) $S_{(AB)} \circ S_{(MN)}$ c) $t_{\vec{BA}} \circ t_{\vec{BC}}$ d) $S_{(AC)} \circ S_{(MN)}$

1. Problem Description

Given a triangle $ABC$ which is right-angled at $A$. $M$ and $N$ are the midpoints of segments $BC$ and $AC$ respectively.

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

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

a) Show that $M$ is the midpoint of segment $AD$.

b) What is the nature of quadrilateral $ABDC$?

3) Characterize:

a) $S_M \circ S_N$

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

c) $t_{\vec{BA}} \circ t_{\vec{BC}}$

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

2. Solution Steps

1) Since $N$ is the midpoint of $AC$, $AN = NC$. Since $MN$ is the line segment connecting the midpoints of $BC$ and $AC$, $MN$ is parallel to $AB$. The symmetry $S_{(MN)}$ maps $C$ to $A$ if and only if $MN$ is the perpendicular bisector of $AC$. Since $ABC$ is a right triangle at $A$, $AB$ is perpendicular to $AC$. Since $MN$ is parallel to $AB$, $MN$ is perpendicular to $AC$. Therefore $MN$ is the perpendicular bisector of $AC$, and thus the symmetry $S_{(MN)}$ maps $C$ to $A$.

2) a) Since $D$ is the image of $B$ under the symmetry $S_{(MN)}$, $MN$ is the perpendicular bisector of $BD$.

Since $M$ is the midpoint of $BC$, $BM = MC$. Also, $MN$ is parallel to $AB$. Let $E$ be the intersection of $MN$ and $BD$. Then $BE = ED$. Since $MN$ is parallel to $AB$, by the midpoint theorem in triangle $CAD$, since $N$ is the midpoint of $AC$, then $MN$ must intersect $AD$ at its midpoint. Also, $MN$ is parallel to $AB$. Let $M'$ be the midpoint of $AD$. Then $MN$ is parallel to $AB$, and $M'$ is the midpoint of $AD$. Consider the coordinates $A(0,0)$, $B(x,0)$, $C(0,y)$. Then $M = (\frac{x}{2}, \frac{y}{2})$ and $N = (0, \frac{y}{2})$. The line $MN$ is $Y=\frac{y}{2}$. Since $D$ is the image of $B$ under $S_{(MN)}$, $D=(x',y')$ with $y'=y-(0-y)=y$. So $D$ is of the form $D=(x',y)$. The midpoint of $BD$ must be on $MN$, $M=(x/2, y/2)$. Since $MN$ is the axis of symmetry, the x coordinates of $B$ and $D$ must be the same. Also, $MN$ must be the perpendicular bisector of $BD$. Since $MN$ is horizontal and $BD$ is vertical, this is true. Hence $D=(x,y)$ is not possible.

Let $D = (a, b)$. Then the midpoint of $BD$ is $(\frac{x+a}{2}, \frac{b}{2})$. This midpoint must lie on the line $MN$, which is $y = \frac{y}{2}$. So $\frac{b}{2} = \frac{y}{2}$, and $b = y$. Thus $D = (a, y)$. The line $BD$ is $x=x$. The slope of $MN$ is $\frac{\frac{y}{2} - \frac{y}{2}}{\frac{x}{2} - 0} = 0$. Thus $MN$ is horizontal. So $B$ and $D$ should be symmetric over the line $y = y/2$. $b = y - y/2 = y/2$, and since $MN$ is a vertical line, $x = 2(0) - x = -x$. Since D is the image of B over the line $y=y/2$, $b'=y-0 = 0$, the coordinates are incorrect.

M is $(\frac{x+0}{2}, \frac{0+y}{2})=(\frac{x}{2},\frac{y}{2})$

Midpoint of $AD$ is $M' = (\frac{x'+0}{2},\frac{y'+0}{2})$. $M' = M = (\frac{x}{2}, \frac{y}{2}) \implies (\frac{x'}{2}, \frac{y'}{2})=(\frac{x}{2},\frac{y}{2}) \implies (x',y')=(x,y)$

Since $MN || AB$ and $MN$ is the perpendicular bisector of $AC$, and $MN$ is the perpendicular bisector of $BD$, then $A$ is the symmetric of $C$ and $B$ is the symmetric of $D$, so we can say that $MN$ is the perpendicular bisector of both AC and BD. Then we can conclude $ADCB$ is a parallelogram, with perpendicular diagonals, and so $ADCB$ is a rhombus. Since $ABC$ is a right triangle with right angle at A, the quadrilateral is not a square.

b) Since $MN$ is the perpendicular bisector of $AC$ and $BD$, and $A$ is the image of $C$ and $B$ is the image of $D$ under the symmetry $S_{MN}$. This means that $ABDC$ is a kite with $AB = AD$ and $CB = CD$. Also, $ABDC$ is a parallelogram, but $AB \neq BD$, $ABDC$ is a rhombus.

3)

a) $S_M \circ S_N$: The composition of two reflections about points M and N is a translation with vector $2 \vec{NM}$.

b) $S_{(AB)} \circ S_{(MN)}$: Since $MN$ is parallel to $AB$, the composition of reflections about parallel lines is a translation.

c) $t_{\vec{BA}} \circ t_{\vec{BC}} = t_{\vec{BA} + \vec{BC}}$: By Chasles's identity, the sum of vectors is $\vec{BD}$ where D is such that ABCD is a parallelogram.

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

3. Final Answer

1) Since N is the midpoint of [AC] and MN is parallel to AB, MN is the perpendicular bisector of [AC]. Therefore, A is the image of C by $S_{(MN)}$.

2) a) M is the midpoint of [AD].

b) ABDC is a rhombus.

3)

a) $S_M \circ S_N$ is the translation $t_{2\vec{NM}}$.

b) $S_{(AB)} \circ S_{(MN)}$ is the translation $t_{2\vec{NM}}$ since $MN || AB$.

c) $t_{\vec{BA}} \circ t_{\vec{BC}} = t_{\vec{BD}}$, where ABDC is a parallelogram.

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