In figure `XY||` seg AC. If `2AX=3BX` and `XY=9`, complete the activity to find the value of AC.

Activity: `2AX=3BX`
`:.(AX)/(BX)=(square)/(square)`
`:.(AX+BX)/(BX)=(square+square)/(square)`.........(By componendo)
`:.(AB)/(BX)=(square)/(square)`.............1
`DeltaBCA~DeltaBYX`.......(`square` test of similarity)
`:.(BA)/(BX)=(AC)/(XY)`
..............(Corresponding sides of similar triangles)
`:.(square)/(square)=(AC)/9`
`:.AC=square` ...........[From 1]

Explanation

`2AX=3BX`
`:.(AX)/(BX)=3/2`
`:.(AX+BX)/(BX)=(3+2)/2` …….(By componendo)
`:.(AB)/(BX)=5/2` ………..1
`DeltaBCA~DeltaBYX`……..(AA test of similarity)
`:.(BA)/(BX)=(AC)/(XY)`
………….(Corresponding sides of similar triangles)
`:.5/2=(AC)/9`
`:.AC=22.5`...........[From 1]