As shown in the figure, in square ABCD, e is a point on CD, extend BC to F, make CF = CE, connect DF, be and DF intersect at g, prove: BG ⊥ DF
It is proved that BC = DC, CE = CF & # 8658; RT △ BCE & # 8773; RT △ DCF
⇒∠CBE=∠CDF
∠BEC=∠DEG
⇒∠DGE=∠BCE=90°
∴BG⊥DF
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