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 select one\

we were given that \\(\angle a\\) is a right angle. that is
(1) \\(m\angle a\\) select one \\(90^{\circ}\\).

furthermore, the contrary statement given above implies:
(2) \\(m\angle b\\) select one \\(90^{\circ}\\) and
(3) \\(m\angle c\\) select one \\(90^{\circ}\\).

using (1)-(3) and addition properties of inequalities, we conclude that \\(m\angle a + m\angle b + m\angle c\\) select one \\(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 is an indirect proof (proof by contradiction) showing that a triangle cannot have more than one right angle.
To prove this, we assume the contrary of the conclusion: "$\triangle ABC$ has two right angles" or "$\triangle ABC$ has at least two angles with measure $90^\circ$".
Thus, the first dropdowns are:

• "$\triangle ABC$ has" <blank>at least two angles</blank>
• "with measure" <blank>\(90^\circ\)</blank> (or <blank>90</blank>)

Express given and contrary conditions mathematically

We are given that $\angle A$ is a right angle, so:
(1) $m\angle A$ <blank>\(=\)</blank> <blank>90</blank>$^\circ$.

The contrary statement implies that at least one other angle is also a right angle, or more generally, since they are positive angle measures in a triangle, if there are at least two right angles, say $\angle B$ and $\angle C$ are also right angles (or at least $\ge 90^\circ$):
(2) $m\angle B$ <blank>\(\ge\)</blank> <blank>90</blank>$^\circ$
(3) $m\angle C$ <blank>\(>\)</blank> <blank>0</blank>$^\circ$ (since any angle in a triangle must be greater than $0^\circ$).
Alternatively, if the contrary assumption is that there are two right angles, say $\angle A$ and $\angle B$, then:
(2) $m\angle B$ <blank>\(=\)</blank> <blank>90</blank>$^\circ$
(3) $m\angle C$ <blank>\(>\)</blank> <blank>0</blank>$^\circ$

Sum the inequalities and find the contradiction

Adding these inequalities:
$m\angle A + m\angle B + m\angle C > 90^\circ + 90^\circ + 0^\circ = 180^\circ$.
Thus, $m\angle A + m\angle B + m\angle C$ <blank>\(>\)</blank> $180^\circ$.

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

Answer:

Proof:

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

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>\(\ge\)</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>\(=
obreak\)</blank> $180^\circ$.