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Segitiga BCD kongruen dengan segitiga CDE
BC = CE
Segitiga ADE siku-siku sama kaki
Sehingga, AE=DE
BD = DE
[tex]AB = 6 AC^{2} = BC^{2} + AB^{2} AC = 6^{2} - 6^{2} AC = 6 \sqrt{2} Cm DE=AE=AC-CE =6 \sqrt{2} - 6 Cm semoga benar ,di periksa dulu kalo ada pilgannya