Question

proof:
first, we assume that this conclusion is false. in other words, we assume that the contrary statement \\\(\triangle abc\\) has select one? with measure select one? is ? \.

we were given that \\(\angle a\\) is a right angle. that is
(1) \\(m\angle a\\) ? \\(^{\circ}\\).

furthermore, the contrary statement given above implies:
(2) \\(m\angle b\\) ? \\(^{\circ}\\) and
(3) \\(m\angle c\\) ? \\(^{\circ}\\).

using (1)-(3) and addition properties of inequalities, we conclude that \\(m\angle a + m\angle b + m\angle c\\) ? \\(180^{\circ}\\).

but this contradicts the select one? which states that \\(m\angle a + m\angle b + m\angle c\\) ? \\(180^{\circ}\\).

Explanation:

Identify the theorem and contrary assumption

The proof aims to show that a triangle cannot have more than one right angle (or obtuse angle) using indirect proof (proof by contradiction).
Let the theorem be: "A triangle can have at most one right angle."
To prove this indirectly, we assume the contrary statement: "$\triangle ABC$ has two right angles" (or "at least two angles with measure $90^\circ$").
Thus, the first sentence is:
"$\triangle ABC$ has <blank>two angles</blank> with measure <blank>\(90^\circ\)</blank> is <blank>true</blank>."

Formulate the mathematical relations

Given that $\angle A$ is a right angle:
(1) \(m\angle A =\) <blank>\(90\)</blank>\(^\circ\).
The contrary statement implies that there is at least one more right angle, say both $\angle B$ and $\angle C$ are also right angles, or at least one of them is:
(2) \(m\angle B =\) <blank>\(90\)</blank>\(^\circ\) and
(3) \(m\angle C =\) <blank>\(90\)</blank>\(^\circ\) (or we assume at least two are $90^\circ$, so \(m\angle B \ge 90^\circ\) or similar, but standard school proofs assume two right angles exactly to show contradiction).
Assuming two right angles, we have:
\(m\angle A = 90^\circ\)
\(m\angle B = 90^\circ\)
\(m\angle C > 0^\circ\)
Adding these:
\(m\angle A + m\angle B + m\angle C > 180^\circ\) (since \(90^\circ + 90^\circ + m\angle C > 180^\circ\)).
Thus, \(m\angle A + m\angle B + m\angle C > 180^\circ\).

Identify the contradiction

This contradicts the <blank>Triangle Angle Sum Theorem</blank>, which states that \(m\angle A + m\angle B + m\angle C =\) <blank>\(180\)</blank>\(^\circ\).

Answer:

First, we assume that this conclusion is false. In other words, we assume that the contrary statement "$\triangle ABC$ has <blank>two angles</blank> with measure <blank>\(90\)</blank> is <blank>true</blank>."

We were given that $\angle A$ is a right angle. That is
(1) \(m\angle A\) <blank>\(=\)</blank> <blank>\(90\)</blank>\(^\circ\).

Furthermore, the contrary statement given above implies:
(2) \(m\angle B\) <blank>\(=\)</blank> <blank>\(90\)</blank>\(^\circ\) and
(3) \(m\angle C\) <blank>\(>\)</blank> <blank>\(0\)</blank>\(^\circ\).

Using (1)-(3) and addition properties of inequalities, we conclude that \(m\angle A + m\angle B + m\angle C\) <blank>\(>\)</blank> \(180^\circ\).

But this contradicts the <blank>Triangle Angle Sum Theorem</blank>, which states that \(m\angle A + m\angle B + m\angle C\) <blank>\(=\)</blank> \(180^\circ\).