Exercise 62: *How many axes of symmetry does an equilateral triangle have? How about an isoceles triangle which is not equilateral?*

$\S 33$) Equilateral triangle having 3 congruent sides can be considered as 3 isoceles triangles by taking each of the 3 vertices as the vertex of the triangle. By $\S 37$ each isoceles triangle has an axis of symmetry. Therefore an equilateral triangle has 3 axes of symmetry.

The non-equilateral isoceles triangle has 1 axis of symmetry per $\S 37$.Exercise 65: *Can a pentagon have an axis of symmetry passing through two (one, none) of its vertices?*

two:

The axis of symmetry requires the pentagon figure to be equally divided about it. Any vertex of a pentagon has a diagonal to two vertices and this excludes the vertices that belong to this vertex's sides. Pentagon $ABCDE$, choose vertex $A$. Diagonals with vertex $A$ are: $\overline{AC}$ and $\overline{AD}$. Sides of vertex $A$ are $\overline{AB}$ and $\overline{AE}$. Other sides of the pentagon are: $\overline{BC}, \overline{CD}, \overline{DE}$. Diagonal $\overline{AC}$ divides the space bounded by the pentagon such that point $B$ lies on one side and points $E$ and $D$ on the other. A similar situation occurs with 1 point on one side of the diagonal and 2 points on the other when the line is drawn to point $D$, $\overline{AD}$. The polygon is not symmetric about the diagonal in either of these cases, the only diagonal cases for a given vertex. Therefore a pentagon cannot have a line of symmetry through two of its vertices.

none:

A pentagon being equally divided about the axis of symmetry requires its vertices to exist equally about or along the line of symmetry. If the axis of symmetry does not intersect any of a pentagon's vertices, half of the vertices must exist on either side of this line. But 5 vertices can not be split equally on either side of the axis of symmetry. Thus the axis of symmetry in a pentagon must intersect a vertex.

Exercise 66: *Two points $A$ and $B$ are given on the same side of a line $MN$. Find a point $C$ on $MN$ such that the line $MN$ would make congruent angles with the sides of the broken line $ACB$.*

Identify the point symmetric about $\overline{MN}$ to point $A$, $A'$. Do the same to identify point $B'$. Connect points $A$ and $B'$ to form $\overline{AB'}$. Connect points $A'$ and $B$ to form $\overline{A'B}$. Label the point of intersection of $\overline{A'B}$ and $\overline{AB'}$, $C$. Figures $ACB$ and $A'CB'$ are symmetric about $\overline{MN}$. Vertical angles $\angle{ACA'}$ and $\angle{BCB'}$ have been formed, and they are congruent by $\S 26$. $\overline{MN}$ bisects these vertical angles. Therefore $\angle{ACM} = \angle{BCN}$.

Exercise 67: *In an isoceles triangle, two medians are congruent, two bisectors are congruent, two altitudes are congruent.*

$\S 37$ In an isoceles triangle the bisector of the angle at the vertex is an axis of symmetry. The congruent angles at either end of the base have an altitude dropped, bisector drawn, and median drawn to their opposing side. The triangle formed with taking one of the base angles as its vertex is congruent to the triangle formed taking the other base angle as the vertex. Therefore there are 2 congruent altitudes, bisectors, and medians in any isoceles triangle.

Exercise 68: *If from the midpoint of each of the congruent sides of an isoceles triangle, the segment perpendicular to this side is erected and continued to its intersection with the other of the congruent sides of the triangle, then these two segments are congruent.*

Exercise 69: *A line perpendicular to the bisector of an angle cuts off congruent segments on its sides.*

$\S 13$ definition of angle. $\S 15$ bisector divides the angle into halves. $\S 37$ the bisector is the axis of symmetry in the angle. Half of the figure on one side of the axis of symmetry can be rotated about the axis of symmetry and identified with the other half in all parts. The perpendicular being symmetric about the bisector, $\S 24$, will also become identified in all points. The point of intersection of the perpendicular and both sides of the angle are now at the same point. Therefore the segments cut-off on the sides of the angle by a perpendicular to the bisector are congruent.

Exercise 70: *An equilateral triangle is equiangular (i.e. all of its angles are congruent).*

An equilateral triangle can be treated as an isoceles triangle, it has 2 congruent sides at every vertex $\S 33$. $\S 35(2)$ In an isoceles triangle, the angles at the base are congruent. Let the angles of the given equilateral triangle by $\angle{1}, \angle{2}, and \angle{3}$. Let $\angle{1}$ be the triangle vertex, $\angle{2}$ and $\angle{3}$ are congruent. Now let $\angle{3}$ be the triangle vertex, $\angle{1}$ and $\angle{2}$ are congruent. But $\angle{2}$ is congruent to $\angle{3}$, so $\angle{3}$ must be congruent to $\angle{1}$. Therefore an equilateral triangle is equiangular.