In fig 1.80, XY ∥ seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC. Figure: triangle ABC with X on side AB and Y on side BC, seg XY ∥ seg AC. Activity: 2AX = 3BX therefore AX/BX = square/square AX + BX/BX = square + square/square …… by componendo AB/BX = square/square …… (I) triangle BCA sim triangle BYX …… □ test of similarity therefore BA/BX = AC/XY …… corresponding sides of similar triangles therefore square/square = AC/9 therefore AC = □ …from (I)

In fig 1.80, XY ∥ seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.

Figure: \(\triangle ABC\) with X on side AB and Y on side BC, seg XY ∥ seg AC.

Activity: 2AX = 3BX \(\therefore \dfrac{AX}{BX} = \dfrac{\square}{\square}\)

\(\dfrac{AX + BX}{BX} = \dfrac{\square + \square}{\square}\) …… by componendo

\(\dfrac{AB}{BX} = \dfrac{\square}{\square}\) …… (I)

\(\triangle BCA \sim \triangle BYX\) …… □ test of similarity

\(\therefore \dfrac{BA}{BX} = \dfrac{AC}{XY}\) …… corresponding sides of similar triangles

\(\therefore \dfrac{\square}{\square} = \dfrac{AC}{9}\) \(\therefore\) AC = □ …from (I)

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Answer: AC = 22.5

2AX = 3BX, so \(\dfrac{AX}{BX} = \dfrac{3}{2}\).

By componendo, \(\dfrac{AX + BX}{BX} = \dfrac{3 + 2}{2}\), so \(\dfrac{AB}{BX} = \dfrac{5}{2}\) …… (I)

In \(\triangle BCA\) and \(\triangle BYX\), \(\angle B\) is common and \(\angle BAC \cong \angle BXY\) (corresponding angles, XY ∥ AC), so \(\triangle BCA \sim \triangle BYX\) by the AA test.

\[\frac{BA}{BX} = \frac{AC}{XY}\] \[\frac{5}{2} = \frac{AC}{9}\] \[AC = \frac{5 \times 9}{2} = 22.5\]

Note: some printed answer keys show 15 here (using \(\frac{5}{3}\)); with AX : BX = 3 : 2 the ratio AB : BX is 5 : 2, so AC = 22.5.

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In fig 1.80, XY ∥ seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.

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