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What's wrong here?

I'm pretty sure I did all the operations correctly so I must be making some kind of conceptual flaw.

(first see the attachment)

[tex] \triangle ABC \sim \triangle BDC [/tex]

[tex] \frac{S_{ABC}}{S_{BDC}}=\frac{AB^2}{BD^2} [/tex]

[tex] \frac{S_{ABC}}{S_{BDC}}=\frac{\frac{AC \times BE}{2}}{\frac{DC \times BE}{2}}=\frac{AC}{DC} \Rightarrow \frac{AC}{DC}=\frac{AB^2}{BD^2} [/tex]

[tex] AB^2=BE^2+AE^2=BC^2-EC^2+(AC-EC)^2=BC^2-EC^2+AC^2-2AC \times EC+EC^2= [/tex]

[tex] =BC^2+AC^2-2AC \times EC [/tex]

[tex] BD^2=BE^2+ED^2=BC^2-EC^2+(EC-DC)^2=BC^2-EC^2+EC^2-2EC \times DC+CD^2= [/tex]

[tex] =BC^2-2EC \times DC+DC^2 [/tex]

[tex] \frac{AC}{DC}=\frac{BC^2+AC^2-2AC \times EC}{BC^2-2EC \times DC+DC^2} [/tex]

[tex] \frac{BC^2-2EC \times DC+DC^2}{DC}=\frac{BC^2+AC^2-2AC \times EC}{AC}[/tex]

[tex] \frac{BC^2}{DC}-2EC+DC=\frac{BC^2}{AC}+AC-2EC [/tex]

[tex] \frac{BC^2}{DC}-AC=\frac{BC^2}{AC}-DC[/tex]

[tex] \frac{BC^2-AC \times DC}{DC}=\frac{BC^2-AC \times DC}{AC} \Rightarrow DC=AC[/tex]

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# A triangle problem

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