One to Two,Isosceles to Isosceles

The $4$ different $\angle A$ possibilities are $\angle A = 108°$ , $\angle A = 90°$ , $\angle A = 25 \frac{5}{7}°$ , and $\angle A = 36°$ , as illustrated below:

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In order to make $2$ triangles out of $1$ triangle from one cut, the cut must go through a vertex and its opposite side. Since $\triangle ABC$ is an isosceles with $\angle B = \angle C$ , so we only need to consider cuts that go through either $\angle A$ or $\angle B$ . In all cases, $AB = BC$ and $\angle B = \angle C$ by isosceles triangle properties.

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Case $A$ : Let the cut intersect $\angle A$ and point $D$ along $BC$ . If the cut results in $2$ isosceles triangles, then there are $9$ cases to consider:

$A1)$ $AB = AD$ and $AD = AC$ . Let $x = \angle C$ . Then $\angle ADC = x$ since $AD = AC$ , $\angle ADB = 180° - x$ since $\angle ADC$ and $\angle ADB$ are linear, $\angle ABD = 180° - x$ since $AB = AD$ , so $x = 180° - x$ since $\angle B = \angle C$ , which means $x = 90°$ , which means $\triangle ACD$ has two right angles, which is not possible.

$A2)$ $AB = AD$ and $AD = CD$ . Let $x = \angle C$ . Then $\angle CAD = x$ since $AD = DC$ , $\angle ADC = 180° - 2x$ by triangle sum of $\triangle ACD$ , $\angle ADB = 2x$ since $\angle ADC$ and $\angle ADB$ are linear, $\angle B = 2x$ since $AB = AD$ , so $x = 2x$ since $\angle B = \angle C$ , which means $x = 0°$ , which is not a possible angle for a triangle.

$A3)$ $AB = AD$ and $AC = CD$ . Let $x = \angle C$ . Then $\angle ADC = 90° - \frac{x}{2}$ by isosceles triangle sum of $\triangle ACD$ , $\angle ADB = 90° + \frac{x}{2}$ since $\angle ADC$ and $\angle ADB$ are linear, $\angle B = 90° + \frac{x}{2}$ since $AB = AD$ , so $x = 90° + \frac{x}{2}$ since $\angle B = \angle C$ , which means $x = 180°$ , which is not a possible angle for a triangle.

$A4)$ $AB = BD$ and $AD = AC$ . By symmetry this is the same as Case $A3$ , which was not possible.

$A5)$ $AB = BD$ and $AD = CD$ . Let $x = \angle C$ . Then $\angle CAD = x$ since $AD = DC$ , $\angle ADC = 180° - 2x$ by triangle sum of $\triangle ACD$ , $\angle ADB = 2x$ since $\angle ADC$ and $\angle ADB$ are linear, $\angle BAD = 2x$ since $AB = BD$ , $\angle B = 180° - 4x$ by triangle sum of $\triangle ACD$ , so $x = 180° - 4x$ since $\angle B = \angle C$ , which means $x = 36°$ , $\angle CAD = x = 36°$ , $\angle BAD = 2x = 2 \cdot 36° = 72°$ , and $\angle A = \angle BAD + \angle CAD = 72°+ 36° = \boxed{108°}$ .

$A6)$ $AB = BD$ and $AC = CD$ . Let $x = \angle C$ . Then $\angle ADC = 90° - \frac{x}{2}$ by isosceles triangle sum of $\triangle ACD$ , $\angle ADB = 90° + \frac{x}{2}$ since $\angle ADC$ and $\angle ADB$ are linear, $\angle BAD = 90° + \frac{x}{2}$ since $AB = AD$ , $\angle B = -x$ by triangle sum of $\triangle ABD$ , so $x = -x$ since $\angle B = \angle C$ , which means $x = 0°$ , which is not a possible angle for a triangle.

$A7)$ $AD = BD$ and $AD = AC$ . By symmetry this is the same as Case $A2$ , which was not possible.

$A8)$ $AD = BD$ and $AD = CD$ . Let $x = \angle C$ . Then $\angle CAD = x$ since $AD = DC$ , $\angle ADC = 180° - 2x$ by triangle sum of $\triangle ACD$ , $\angle ADB = 2x$ since $\angle ADC$ and $\angle ADB$ are linear, $\angle B = \angle BAD = 90° - x$ by isosceles triangle sum of $\triangle ABD$ , so $x = 90° - x$ since $\angle B = \angle C$ , which means $x = 45°$ , $\angle CAD = x = 45°$ , $\angle BAD = 90° - x = 90° - 45° = 45°$ , and $\angle A = \angle BAD + \angle CAD = 45°+ 45° = \boxed{90°}$ .

$A9)$ $AD = BD$ and $AC = CD$ . By symmetry this is the same as Case $A5$ , which would give $\angle A = 108°$ again.

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Case $B$ : Let the cut intersect $\angle B$ and point $D$ along $AC$ . If the cut results in $2$ isosceles triangles, then there are $9$ cases to consider:

$B1)$ $AB = BD$ and $BC = BD$ . Let $x = \angle C$ . Then $\angle BDC = x$ since $BC = BD$ , $\angle ADB = 180° - x$ since $\angle BDC$ and $\angle ADB$ are linear, $\angle A = 180° - x$ since $AB = BD$ , $\angle CBD = 180° - 2x$ by triangle sum of $\triangle BCD$ , $\angle ABD = 2x - 180°$ by triangle sum of $\triangle ABD$ , $\angle B = 0°$ as the sum of $\angle CBD$ and $\angle BCD$ , which is not a possible angle for a triangle.

$B2)$ $AB = BD$ and $BD = CD$ . Let $x = \angle C$ . Then $\angle CBD = x$ since $BD = CD$ , $\angle BDC = 180° - 2x$ by triangle sum of $\triangle BCD$ , $\angle ADB = 2x$ since $\angle BDC$ and $\angle ADB$ are linear, $\angle A = 2x$ since $AB = BD$ , $\angle ABD = 180° - 4x$ by triangle sum of $\triangle ABD$ , $\angle B = 180° - 3x$ as the sum of $\angle CBD$ and $\angle BCD$ , so $x = 180° - 3x$ since $\angle B = \angle C$ , which means $x = 45°$ , $\angle ADB = 2x = 2 \cdot 45° = 90°$ , and $\angle A = 2x = 2 \cdot 45° = 90°$ , which means $\triangle ABD$ has two right angles, which is not possible.

$B3)$ $AB = BD$ and $BC = CD$ . Let $x = \angle C$ . Then $\angle CBD = \angle CDB = 90° - \frac{x}{2}$ by isosceles triangle sum of $\triangle BCD$ , $\angle ADB = 90° + \frac{x}{2}$ since $\angle BDC$ and $\angle ADB$ are linear, $\angle A = 90° + \frac{x}{2}$ since $AB = BD$ , $\angle ABD = -x$ by triangle sum of $\triangle ABD$ , $\angle B = 90° - \frac{3x}{2}$ as the sum of $\angle CBD$ and $\angle BCD$ , so $x = 90° - \frac{3x}{2}$ since $\angle B = \angle C$ , which means $x = 72°$ and $\angle ABD = -x = -72°$ , which is not a possible angle for a triangle.

$B4)$ $AD = BD$ and $BC = BD$ . Let $x = \angle C$ . Then $\angle BDC = x$ since $BC = BD$ , $\angle ADB = 180° - x$ since $\angle BDC$ and $\angle ADB$ are linear, $\angle CBD = 180° - 2x$ by triangle sum of $\triangle BCD$ , $\angle A = \angle ABD = \frac{x}{2}$ by isosceles triangle sum of $\triangle ABD$ , $\angle B = 180° - \frac{3x}{2}$ as the sum of $\angle CBD$ and $\angle BCD$ , so $x = 180° - \frac{3x}{2}$ since $\angle B = \angle C$ , which means $x = 72°$ and $\angle A = \frac{x}{2} = \frac{72°}{2} = \boxed{36°}$ .

$B5)$ $AD = BD$ and $BD = CD$ . Let $x = \angle C$ . Then $\angle CBD = x$ since $BD = CD$ , $\angle BDC = 180° - 2x$ by triangle sum of $\triangle BCD$ , $\angle ADB = 2x$ since $\angle BDC$ and $\angle ADB$ are linear, $\angle A = \angle ABD = 90° - x$ by isosceles triangle sum of $\triangle ABD$ , $\angle B = 90°$ as the sum of $\angle CBD$ and $\angle BCD$ , so $x = 90°$ since $\angle B = \angle C$ , which means $\angle B = \angle C = 90°$ , which means $\triangle ABC$ has two right angles, which is not possible.

$B6)$ $AD = BD$ and $BC = CD$ . Let $x = \angle C$ . Then $\angle CBD = \angle CDB = 90° - \frac{x}{2}$ by isosceles triangle sum of $\triangle BCD$ , $\angle ADB = 90° + \frac{x}{2}$ since $\angle BDC$ and $\angle ADB$ are linear, $\angle A = \angle ABD = 45° - \frac{x}{4}$ by isosceles triangle sum of $\triangle ABD$ , $\angle B = 135° - \frac{3x}{4}$ as the sum of $\angle CBD$ and $\angle BCD$ , so $x = 135° - \frac{3x}{4}$ since $\angle B = \angle C$ , which means $x = 77\frac{1}{7}°$ and $\angle A = 45° - \frac{x}{4} = 45° - \frac{77\frac{1}{7}°}{4} = \boxed{25\frac{5}{7}°}$ .

$B7)$ $AB = AD$ and $BC = BD$ . Then $AC = AB = AD$ , and $CD = AC - AD = 0$ , which is not a possible length for a triangle.

$B8)$ $AB = AD$ and $BD = CD$ . Then $AC = AB = AD$ , and $CD = AC - AD = 0$ , which is not a possible length for a triangle.

$B9)$ $AB = AD$ and $BC = CD$ . Then $AC = AB = AD$ , and $CD = AC - AD = 0$ , which is not a possible length for a triangle.

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Therefore, the only different $\angle A$ possibilities are $108°$ , $90°$ , $25 \frac{5}{7}°$ , and $36°$ .

Label the angles as follows:

The angles satisfy the equations $\begin{cases} a + b = q \\ c + d = p \\ a + b + c + d = p + q = 180^\circ \end{cases}$

Now consider the different cases, based on which angles are equal in the three triangles. In each triangle there are potentially three angles that could be the vertex angle, making for 27 cases in principle; but several of them can be ruled out, because

• $a$ or $d$ cannot be the vertex angle in both the large and a small triangle;

  • if $b+c$ is the vertex angle in the large triangle, neither $b$ nor $c$ can be the vertex angle in the smaller triangles;

  • if $d$ is the vertex angle in the large triangle, $p$ cannot be the vertex angle in a small triangle; the same holds for $a$ and $q$ ;

  • the symmetry between $(a,b,p)$ and $(c,d,q)$ allows us to rule out some more cases.

That leaves the following cases

$\begin{array}{cccc} \text{case} & \text{large triangle} & \text{small triangle 1} & \text{small triangle 2} \\ \hline (1) & a = d & a = b & c = d \\ (2) & a = d & a = b & c = q \\ (3) & a = d & b = p & c = q \\ (4) & a = b+c & a = p & c = d \\ (5) & a = b+c & a = p & d = q \\ (6) & a = b+c & b = p & c = d \\ (7) & a = b+c & b = p & d = q \\ \hline \end{array}$

For each case, we substitute the given values into the three equations listed at the beginning. We obtain:

• Case (1): $a = b = c = d$ , $4a = 180^\circ$ , $a = 45^\circ$ , vertex angle $b + c = \boxed{90^\circ}$ .

• Case (2): $a = b = d$ , $3a + c = 180^\circ$ , (2a = c), $5a = 180^\circ$ , $a = 36^\circ$ , vertex angle $b + c = 3a = \boxed{108^\circ}$ .

• Case (3): $a = d$ , $a + b = c$ , $c + a = b$ , which gives $a = 0$ , which is impossible.

• Case (4): $b = c = d$ , $a = 2c$ , $5c = 180^\circ$ , $c = 36^\circ$ , vertex angle $d = c = \boxed{36^\circ}$ .

• Case (5): $b = d = a+b$ , so that $a = 0$ , which is impossible.

• Case (6): $a = 3c$ , $b = p = 2c$ , $q = 5c$ , $7c = 180^\circ$ , vertex angle $d = c = \boxed{25\tfrac57^\circ}$ .

• Case (7): $a = p+c = 2c+d = 2c+a+b$ , so that $2c + b = 0$ , which is impossible.

Thus we see that there are $\boxed{\text{four}}$ possible vertex angles for the large triangle.