Question

consider the two triangles.
to prove that the triangles are similar by the sas similarity theorem, it needs to be shown that
\\(\frac{ac}{gi} = \frac{hi}{bc}\\)\\.
\\(\angle c \cong \angle c\\).
\\(\angle c \cong \angle g\\).
\\(\frac{ac}{gi} = \frac{bc}{hi}\\).

Explanation:

Step1: Recall SAS Similarity Theorem

The SAS (Side - Angle - Side) similarity theorem states that if two sides of one triangle are proportional to two sides of another triangle and the included angles are congruent, then the triangles are similar.

Step2: Identify the sides and included angle

In triangle \(ABC\) and triangle \(HIG\) (assuming the right angles are \(\angle C\) and \(\angle I\)), the sides adjacent to the right angles:
- In \(\triangle ABC\), \(AC = 15\), \(BC=20\)
- In \(\triangle HIG\), \(HI = 12\), \(GI = 9\)

We need to check the ratios of the corresponding sides and the congruence of the included angles. The included angle for the sides \(AC,BC\) in \(\triangle ABC\) is \(\angle C\), and for the sides \(HI,GI\) in \(\triangle HIG\) is \(\angle I\)? Wait, no, let's re - examine. Wait, actually, let's check the ratios:

\(\frac{AC}{GI}=\frac{15}{9}=\frac{5}{3}\) and \(\frac{BC}{HI}=\frac{20}{12}=\frac{5}{3}\)

Also, the included angle: \(\angle C\) (in \(\triangle ABC\)) and \(\angle I\)? Wait, no, the right angles: \(\angle C\) and \(\angle I\) are right angles? Wait, the diagram shows \(\angle C\) and \(\angle I\) as right angles? Wait, no, in the first triangle, right angle at \(C\), in the second triangle, right angle at \(I\). Wait, maybe I misread. Wait, the options have \(\angle C\cong\angle I\)? No, the options have \(\frac{AC}{GI}=\frac{BC}{HI}\) and \(\angle C\cong\angle I\)? Wait, the options given:

Let's check each option:

Option 1: \(\frac{AC}{GI}=\frac{HI}{BC}\). \(\frac{15}{9}=\frac{5}{3}\), \(\frac{HI}{BC}=\frac{12}{20}=\frac{3}{5}\). Not equal.

Option 2: \(\angle C\cong\angle C\). Doesn't make sense, probably a typo, maybe \(\angle C\cong\angle I\)? But the option is \(\angle C\cong\angle C\), which is wrong.

Option 3: \(\angle C\cong\angle G\). No, \(\angle C\) is a right angle, \(\angle G\) is not a right angle (since \(GI = 9\), \(HI=12\), so \(\angle I\) is right angle).

Option 4: \(\frac{AC}{GI}=\frac{BC}{HI}\). We calculated \(\frac{AC}{GI}=\frac{15}{9}=\frac{5}{3}\) and \(\frac{BC}{HI}=\frac{20}{12}=\frac{5}{3}\), and the included angle: since \(\angle C\) (right angle) and \(\angle I\) (right angle), so \(\angle C\cong\angle I\), but the option for the angle congruence is not there, but among the given options, the ratio \(\frac{AC}{GI}=\frac{BC}{HI}\) is correct, and also, if we consider the included angle, but the key is the SAS similarity: two sides proportional and included angle congruent. The ratio \(\frac{AC}{GI}=\frac{BC}{HI}\) and the included angle \(\angle C\cong\angle I\) (both right angles). But among the options, the ratio \(\frac{AC}{GI}=\frac{BC}{HI}\) is \(\frac{15}{9}=\frac{20}{12}\) (simplify: \(\frac{5}{3}=\frac{5}{3}\)) and the included angle (the angle between \(AC\) and \(BC\) is \(\angle C\), and between \(GI\) and \(HI\) is \(\angle I\), but if we assume that \(\angle C\cong\angle I\) (both right angles), and \(\frac{AC}{GI}=\frac{BC}{HI}\), then by SAS similarity. The option \(\frac{AC}{GI}=\frac{BC}{HI}\) is correct.

Wait, the correct ratio is \(\frac{AC}{GI}=\frac{BC}{HI}\) (since \(\frac{15}{9}=\frac{20}{12}=\frac{5}{3}\)) and the included angle \(\angle C\cong\angle I\) (both right angles). Among the given options, the option \(\frac{AC}{GI}=\frac{BC}{HI}\) is \(\frac{AC}{GI}=\frac{BC}{HI}\), which is \(\frac{15}{9}=\frac{20}{12}\), and also, the included angle (the angle between \(AC\) and \(BC\) is \(\angle C\), and between \(GI\) and \(HI\) is \(\angle I\), but if we consider the right angles, \(\angle C\) and \(\angle I\) are right angles, so \(\angle C\cong\angle I\). But the option with the ratio \(\frac{AC}{GI}=\frac{BC}{HI}\) is correct.

Answer:

\(\boldsymbol{\frac{AC}{GI}=\frac{BC}{HI}}\) (the fourth option: \(\frac{AC}{GI}=\frac{BC}{HI}\))