In the adjoining figure, seg `XY||` seg `AC`, IF `3AX=2BX` and `XY=9` then find the length of `AC`.
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`3AX=2BX`
`:.(AX)/(BX)=(2)/(3)`
By componendo, we get
`(AX+BX)/(BX)=(2+3)/(3)`
`(AB)/(BX)=(5)/(3)`
`:.` by invertendo, we get
`(BX)/(AB)=(3)/(5)`……`(1)`
In `DeltaBXY` and `DeltaBAC` ,
`/_BXY~=/_BAC`…….(Corresponding angles)
`/_XBY~=/_ABC`…….(Common angle)
`:.DeltaBXY~DeltaBAC`……(“AA” test of similarity)
`:.(BX)/(AB)=(XY)/(AC)`……(Corresponding sides of similar triangle)
`:.(3)/(5)=(9)/(AC)`……..[From `(1)`]
`:.3xxAC=9xx5`
`:.AC=(9xx5)/(3)`
`AC=15`