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Question
In the given fig, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find `"AX"/"XY"`.
Solution
In Δ ABY,
ray BX bisects ∠ABY ...(Given)
∴ by the theorem of an angle bisector of a triangle,
`("AB")/("BY") = ("AX")/("XY")` ...(1)
In Δ ACY,
ray CX bisects ∠ACY ...(Given)
∴ by the theorem of an angle bisector of a triangle,
`("AC")/("CY") = ("AX")/("XY")` ...(2)
`("AB")/("BY") = ("AC")/("CY") = ("AX")/("XY")` ...[From (1) and (2)]
∴ `5/("BY") = 4/("CY") = ("AX")/("XY")`
By theorem on equal ratios,
`(5 + 4)/("BY" + "CY") = ("AX")/("XY")`
∴ `9/("BC") = ("AX")/("XY")` ...(B - Y - C)
∴ `9/6 = ("AX")/("XY")`
∴ `("AX")/("XY") = 3/2`
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