Question

show that $a^2 + b^2 = c^2$, so $\angle a \cong \angle r$. this means $\triangle abc \cong \triangle rts$, and so $\angle c$ is a right angle and so $\triangle abc$ is a right triangle.
show that $c = x$, so $\triangle abc \cong \triangle rts$ and so $\angle c \cong \angle s$. this means $\angle c$ is a right angle, and so $\triangle abc$ is a right triangle
show that $c = x$, so $\triangle abc \sim \triangle rts$ and so $\angle c \sim \angle s$. this means $\angle c$ is a right angle, and so $\triangle abc$ is a right triangle.
show that $c \
eq x$, so $\triangle abc \cong \triangle rts$ and so $\angle c \cong \angle s$. this means $\angle c$ is a right angle, and so $\triangle abc$ is a right triangle.

Explanation:

To determine the correct option, we analyze the triangles:

1. Triangles \( \triangle ABC \) (red) and \( \triangle RTS \) (blue) have two sides equal: \( AC = RS = a \) and \( BC = TS = b \). If \( c = x \) (hypotenuses equal), by SSS (Side - Side - Side) congruence criterion, \( \triangle ABC \cong \triangle RTS \).
2. In \( \triangle RTS \), \( \angle S \) is a right angle (since it's a right - angled triangle with legs \( a,b \) and hypotenuse \( x \)). If \( \triangle ABC \cong \triangle RTS \), then \( \angle C\cong\angle S \), so \( \angle C \) is a right angle, making \( \triangle ABC \) a right triangle.
3. Analyzing other options:

• The first option's reasoning about \( a^{2}+b^{2}=c^{2} \) and \( \angle A\cong\angle R \) is incorrect for proving congruence leading to \( \angle C \) being right.
• The third option uses similarity (\( \sim \)) instead of congruence (\( \cong \)). Similarity doesn't guarantee equal angles in the same way for proving a right angle here.
• The fourth option states \( c

eq x \), which would not satisfy SSS congruence.

Answer:

Show that \( c = x \), so \( \triangle ABC\cong\triangle RTS \) and so \( \angle C\cong\angle S \). This means \( \angle C \) is a right angle, and so \( \triangle ABC \) is a right triangle.