$B'$ and $A'C'$ should be $B_1$ and $A_1C_1$ respectively

Exercise 96: *Each leg of a right triangle is smaller than the hypotenuse.*

Each leg in a right triangle is not opposite the right angle, $\S 33$) right triangle definition. Since the right angle has an exterior angle that is also right, $\S 22$, the remaining interior angles must be < $d$, $\S 43$. By $\S 45(2)$ the side opposite to a greater angle is greater. Therefore the legs must be smaller than the hypotenuse because they are opposite acute angles.

Exercise 97: *A right triangle can have at most one axis of symmetry.*

$\S 37$) Axis of symmetry - figure is symmetric to itself about a line. Symmetric - two points are situated on the opposite sides of a line, on the same perpendicular to this line, and the same distance from the foot of the perpendicular.

The axis of symmetry must intersect the vertex of one of the angles so sub-triangles are created by it. Otherwise the axis of symmetry would intersect at two new points on the triangle, creating a quadrilateral on one of its sides which can never be congruent to the triangle formed on its other side.

The axis of symmetry divides the right triangle into two triangles, one on either side of the axis. Either of these triangles can be rotated about the axis of symmetry until it falls back onto the plane and is superimposed onto the other triangle. If the axis of symmetry is through either of the acute angles, the hypotenuse will lie on one side of the axis. By #$96$ the hypotenuse is larger than the legs and will not be congruent to a part of the figure on the other side of such an axis of symmetry.

The axis of symmetry must then cut the right angle as a bisector to keep the two angles on either side symmetric about it, as midpoint to the hypotenuse to maintain symmetry of the hypotenuse parts about it, and the legs which exist on either side of the axis must be congruent. Right triangle, congruent legs, axis of symmetry is bisector of right angle conditions met, $\S 40(1)$ SAS congruence of sub-triangles on either side of axis of symmetry: $leg = leg$, $\frac{d}{2} = \frac{d}{2}$, $bisector = bisector$.

Exercise 98: *At most two congruent slants to a given line can be drawn from a given point.*

From this same point drop a perpendicular to the given line. The foot of the slants must exist on either side of the foot of the perpendicular. $\S 53(1)$ if two slants are congruent, then thir feet are the same distance away from the foot of the perpendicular. On one side of the perpendicular select a point on the given line for the feet of the congruent slants. All of the congruent slants on this side of the perpendicular must be dropped from the given point to the selected point. $\S 4$) For every two points in space there exists exactly one line. Therefore only one such slant can exist on this side of the perpendicular.

A point the same distance from the foot of the perpendicular can be selected along the given line on other side of the perpendicular. As shown above only one such slant can exist on that side. There are only two congruent slants from the given point to the given line.

Exercise 99: *Two isoceles triangles with a common vertex and congruent lateral sides cannot fit one inside the other.*

I must misunderstand the problem. It appears to be a straight forward case of $\S 40(1)$ SAS, the two isoceles triangles are congruent and neither is smaller or larger than the other.

Exercise 100: *The bisector of an angle is its axis of symmetry.*

$\S 37$) Axis of symmetry - figure is symmetric to itself about a line. Symmetric - two points are situated on the opposite sides of a line, on the same perpendicular to this line, and the same distance from the foot of the perpendicular.

If the exercise is about "angle", $\S 15$ defines bisector as the ray dividing a given angle into halves. I.e. the plane can be folded along the bistor, the bisector ray position will be unchanged and the two angle sides will have become identified with each other. Thus an angle bisector is an axis of symmetry.

If the exercise is about a "triangle", $\S 34$) triangle bisector halves the vertex angle. The bisector cuts the base into two segments that must be congruent and along the same perpendicular to the bisector for it to be the axis of symmetry. This implies that the bisector is the median and the altitude. Under these conditions in the bisected triangle, two sub-triangles exist about the bisector where each of these share congruent halves of the original vertex, a common side (the bisector), and conguent right angles at the foot of the altitude. $\S 40(2)$ ASA - the sub-triangles are congruent, so their hypotenuses (laterals of the original triangle) are congruent as well. The bisector is an axis of symmetry for the triangle when it is also the median and altitude, also know as isoceles triangle.

Exercise 101: *A triangle is isoceles if two of its altitudes are congruent.*

In $\triangle ABC$ drop altitude $AD$ from vertex $A$ and drop altitude $CE$ from vertex $C$. Let these altitudes be congruent. Two right triangles formed by these altitudes are $\triangle ADC$ and $\triangle CEA$, which share $AC$ as their hypotenuse. By $55(2)$ if the hypotenuse and a leg of one right triangle are congruent respectively to the hypotenuse and a leg of another right triangle, the triangles are congruent. It follows that $\angle A = \angle C$.

$\S 45(1)$ In any triangle the sides opposite congruent angles are congruent; $BC = AB$. $\S 33$) Isoceles triangle by definition has 2 congruent sides.

Exercise 102: *A median in a triangle is equidistant from the two vertices not lying on it.*

In the case that the median is also the altitude, the base and median are perpendicular. The summands of the base as divided by the foot of the median are congruent and by $\S 51$ these perpendiculars are the shortest segments from the vertices at their ends to the median.

Given $\triangle ABC$ in which the median does not form a right angle with the base. Constructing a perpendicular from a vertex at either end of the base to the median forms a right triangle, with half the base as the hypotenuse. The angles at both ends of the hypotenuse in this right triangle must be acute by the properties of the right triangle and $\S 42$. If the angle made by the median and one half of the base on the interior of $\triangle ABC$ is obtuse, the median must be extended past the base and the perpendicular constructed exterior to the original triangle. Note that for any median that is not an altitude this will be the case for one of the perpendiculars being constructed, according to supplementary angles $\S 22$.

The two right triangles that have been formed by these perpendiculars have hypotenuses that are continuations of one another (the base of $\triangle ABC$) and their sides along the median are continuations of one another. By $\S 26$ these right triangles have acute angles that are vertical and congruent. $\S 55(1)$ These right triangles have congruent hypotenuses and an acute angle, so they are congruent. The perpendiculars of the right triangles are congruent then, $\S 51$ these perpendiculars are the distance from the vertices at the ends of the base of $\triangle ABC$ to the median and are congruent.

Exercise 103: *A line and a circle can have at most two common points.*

$\S 9$) Circle - a curved line all of whose points are the same distance from a select point.

Let there be a circle with a straight line intersecting it. For every point of intersection (common) there must be a radius from the center of the circle to it. These radii are congruent. #$98$ proved that at most two congruent slants to a given line exist from the center of the circle. $\S 51$) The perpendicular dropped from the center of the circle to the given line is shorter than any slant from the center point to this line. Thus there are at most two common points.

This also shows that when there is a common point that is on the perpendicular from the circle center to the given line, it is the only common point as the slants must be larger and not congruent to the circle radius.