At a given vertex, an exterior angle and an interior angle of a triangle form a linear pair, making them supplementary - that is, their measures total $\displaystyle 180^{\circ }$. The measures of two interior angles can be calculated by subtracting the exterior angle measures from $\displaystyle 180^{\circ }$:

$\displaystyle 180^{\circ } - 145^{\circ } = 35^{\circ }$

$\displaystyle 180^{\circ } - 125^{\circ } = 55^{\circ }$

The triangle has two interior angles of measures $\displaystyle 35^{\circ }$ and $\displaystyle 55^{\circ }$. The sum of these measures is $\displaystyle 35^{\circ }+ 55 ^{\circ } = 90 ^{\circ }$, thereby making them complementary. A triangle with two complementary acute angles is a right triangle.

The sum of the measures of the angles of a triangle is $\displaystyle 180^{\circ }$. The sum of the three given angle measures is 

$\displaystyle 50^{\circ }+ 60 ^{\circ }+ 70^{\circ } = 180^{\circ }$.

This makes the triangle possible.

A triangle with three congruent angles is an equiangular - and equilateral - triangle; such an angle must have three angles that measure $\displaystyle 60^{\circ }$. 

The perimeter of $\displaystyle \bigtriangleup ABC$ is the sum of the lengths of its sides - that is,

$\displaystyle AB + BC + AC = P$

The perimeter is 40, so set $\displaystyle AB = 12 , BC = 14, P = 40$, and solve for $\displaystyle AC$:

$\displaystyle AB + BC + AC = P$

$\displaystyle 12+14 + AC = 40$

$\displaystyle 26 + AC = 40$

Subtract 26 from both sides:

$\displaystyle 26 + AC - 26 = 40 - 26$

$\displaystyle AC= 14$

$\displaystyle \overline{BC}\cong \overline{AC}$, so by the Isosceles Triangle Theorem, their opposite angles are congruent - that is,

$\displaystyle \angle A \cong \angle B$.

The referenced triangle is below:

In an equilateral triangle, the median from $\displaystyle A$ - the segment from $\displaystyle A$ to $\displaystyle M$, the midpoint of the opposite side $\displaystyle \overline{BC}$ - is also the bisector of the angle $\displaystyle \angle BAC$, so 

$\displaystyle m \angle BAM = \frac{1}{2} \cdot \angle BAC$

Each interior angle of an equilateral triangle, including $\displaystyle \angle BAC$, measures $\displaystyle 60^{\circ }$, so substitute and evaluate:

$\displaystyle m \angle BAM = \frac{1}{2} \cdot 60^{\circ } = 30 ^{\circ }$.

The referenced figure is below. Note that $\displaystyle m \angle C = 60 ^{\circ }$, as is the case with all of the interior angles of an equilateral triangle.

The interior angles of an equilateral triangle each measure $\displaystyle 60^{\circ }$. An exterior angle of a triangle has as its degree measure the sum of its remote interior angles; specifically, 

$\displaystyle m \angle ADB = m\angle C + m \angle DAC$

Substitute the known angle measures, and solve:

$\displaystyle 80^{\circ }= 60^{\circ } + m \angle DAC$

$\displaystyle 80^{\circ }- 60^{\circ } = 60^{\circ } + m \angle DAC - 60^{\circ }$

$\displaystyle 20^{\circ } = m \angle DA C$

To find the length of the hypotenuse, apply the Pythagorean Theorem: $\displaystyle a^2+b^2=c^2$, where $\displaystyle c=$ the length of the hypotenuse. 

Thus, the solution is:

$\displaystyle 4^2+10^2=c^2$

$\displaystyle 16+100=c^2$

$\displaystyle c^2=116$

$\displaystyle c=\sqrt{116}=\sqrt{4\times 29}=\sqrt{4} \sqrt{29}=2\sqrt{29}$

To find the length of the hypotenuse, apply the Pythagorean Theorem: $\displaystyle a^2+b^2=c^2$, where $\displaystyle c=$ the length of the hypotenuse. 

Thus, the solution is:

$\displaystyle 9.5^2+9.5^2=c^2$

$\displaystyle c^2=90.25+90.25=180.5$

$\displaystyle c=\sqrt{180.5}$

To find the length of the hypotenuse, apply the Pythagorean Theorem: $\displaystyle a^2+b^2=c^2$, where $\displaystyle c=$ the length of the hypotenuse. 

Thus, the solution is:

$\displaystyle 12^2+8^2=c^2$

$\displaystyle c^2=144+64=208$

$\displaystyle c=\sqrt{208}=\sqrt{16\times 13}=\sqrt{16}\sqrt{13}=4\sqrt{13}$

To find the length of the hypotenuse, apply the Pythagorean Theorem: $\displaystyle a^2+b^2=c^2$, where $\displaystyle c=$ the length of the hypotenuse. 

Thus, the solution is:

$\displaystyle 11^2+11^2=c^2$

$\displaystyle c^2=121+121=242$

$\displaystyle c=\sqrt{242}=\sqrt{121\times 2}=\sqrt{121}\sqrt{2}=11\sqrt{2}$