Question

In the following figure, XY \(||\) seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.

Activity: 
2AX = 3BX (Given) 
\[\therefore \frac{AX}{BX} = \frac{3}{\boxed{2}} \ \\ \frac{AX + BX}{BX} = \frac{3 + 2}{2} \quad \text{(by componendo)} \ \\ \frac{BA}{BX} = \frac{5}{2} \quad \dots \text{(I)} [6pt] \\ \text{Now } \triangle BCA \sim \triangle BYX ; (\boxed{\text{AA}} \text{ test of similarity}) [4pt] \\ \therefore \frac{BA}{BX} = \frac{AC}{XY} \quad \text{(corresponding sides of similar triangles)} [4pt] \\ \frac{5}{2} = \frac{AC}{9} \quad \text{from (I)} [4pt] \\ \therefore AC = \boxed{22.5}\]
 

Hint:

When dealing with similar triangles, be very careful to match the corresponding vertices correctly. The ratio of sides must be between corresponding sides. For example, if \(\triangle ABC \sim \triangle DEF\), then \(\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}\).

Answer 1

Step 1: Understanding the Concept:
This activity uses the properties of similar triangles. Since XY \(||\) AC, \(\triangle\)BYX is similar to \(\triangle\)BCA. The ratio of their corresponding sides is equal. The given relation between AX and BX allows us to find the ratio of the sides BA and BX.

Step 2: Key Formula or Approach:
1. AA Similarity Test: If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar.
2. If two triangles are similar, the ratio of their corresponding sides is equal.

Step 3: Detailed Explanation:
Here is the completed activity with the blanks filled in.

2AX = 3BX (Given)
\(\therefore \frac{AX}{BX} = \frac{3}{\boxed{2}}\)
\(\frac{AX + BX}{BX} = \frac{3+2}{2}\) (by componendo)
Since AX + BX = BA,
\(\frac{BA}{BX} = \frac{5}{2}\) \(\dots\) (I)

Now consider \(\triangle\) BCA and \(\triangle\) BYX.
Since XY \(||\) AC and BC is a transversal, \(\angle BYX \cong \angle BCA\) (corresponding angles).
Also, \(\angle XBY \cong \angle ABC\) (common angle).
Therefore, by AA similarity test, \(\triangle\) BYX \(\sim\) \(\triangle\) BCA.
The activity has the order of vertices as \(\triangle\) BCA \(\sim\) \(\triangle\) BYX.

Now \(\triangle\) BCA \(\sim\) \(\triangle\) BYX (\(\boxed{\text{AA}}\) test of similarity)
\(\therefore \frac{BA}{BY} = \frac{CA}{YX} = \frac{BC}{BX}\). The activity uses \(\frac{BA}{BX} = \frac{AC}{XY}\) which implies a similarity correspondence of \(\triangle BAC \sim \triangle BXY\).
Let's check this. \(\angle XBY \cong \angle ABC\) (Common angle). \(\angle BXY \cong \angle BAC\) (Corresponding angles). So, \(\triangle BXY \sim \triangle BAC\) is correct. The ratio of corresponding sides would be \(\frac{BX}{BA} = \frac{BY}{BC} = \frac{XY}{AC}\).
The activity seems to have a typo in the side ratio correspondence, but we follow its logic. It states:
\(\frac{BA}{BX} = \frac{AC}{XY}\) (corresponding sides of similar triangles)

Substitute the values from (I) and the given value of XY:
\(\frac{5}{2} = \frac{AC}{9}\) from (I)

Now solve for AC:
AC = \(\frac{5 \times 9}{2} = \frac{45}{2} = 22.5\)
\(\therefore\) AC = \(\boxed{22.5}\)

Step 4: Final Answer:
The value of AC is 22.5.