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Sagot :
Answer:
[tex]Here,\\Since~BC[/tex]║[tex]DE,\\[/tex]
[tex]<ABC = <ADE and <ACB=<AED\\Therefore,~ABC~ and~ ADE ~are~ similar~ triangles.\\Now,\\\\\frac{AC}{AE} = \frac{AB}{AD}\\[/tex]
[Corresponding Sides of similar triangles are proportional.]
[tex]or, \frac{y}{2y} = \frac{3}{3y} \\or, \frac{1}{2} = \frac{1}{y}\\or, y=2[/tex]
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