Acute plane triangle with two sides coinciding with a right triangle Below are the two questions:
1) If $T$ is a plane triangle with $x, y < z$ such that $x^{2} + y^{2} > z^{2}$
as side lengths, is $T$ necessarily acute?
2) Is an acute plane triangle $T$ such that 
$T$ has all sides of integer lengths
$T$ has exactly two sides coinciding with a right plane triangle but not necessarily of equal length
necessarily equilateral?
 A: The first first question can be answered via the Law of Cosines. Given a triangle with side lengths $a,b,c$ and angle $\gamma$ opposite to side $c$, the Cosine Law states:
$$c^2=a^2+b^2-2ab\cos{\gamma}.$$
The identity can be arranged to isolate the cosine term:
$$\cos\gamma=\frac{a^2+b^2-c^2}{2ab}.$$
Thus, if we suppose that $a^2+b^2>c^2$, then it follows that the cosine of the angle must be greater than zero too. That means the measure of the angle must be less than $90^\circ$ :
$$\cos\gamma>0\\
\implies 0<\gamma<\frac{\pi}{2}.$$
Note that if $c$ is the longest side, then the other two relevant inequalities $b^2+c^2>a^2$ and $c^2+a^2>b^2$ will hold true automatically for any triangle. Hence, by adding the assumption $a^2+b^2>c^2$, all internal angles of the triangle must be less than $90^\circ$.

As for the second question, the answer is obviously no. All you need to prove it is a counter-example. Here's a 3-4-5 right plane triangle (it has integer side lengths), and as you can see, it wasn't hard to inscribe a non-equilateral triangle:

