25. (10 分)如图,在$Rt△ ACB$中,$∠ ACB = 90^{\circ}$,点$D$是$AB$的中点,点$E$是$CD$的中点,过点$C$作$CF // AB$交$AE$的延长线于点$F$.
(1)求证:$△ ADE ≌ △ FCE$;
(2)若$∠ DCF = 120^{\circ}$,$DE = 2$,求$BC$的长.

(1)求证:$△ ADE ≌ △ FCE$;
(2)若$∠ DCF = 120^{\circ}$,$DE = 2$,求$BC$的长.
答案
25. (1) 证明略. (2) $BC = 4$.
26. (12 分)如图,现有一张边长为$4$的正方形纸片$ABCD$,点$P$为正方形$AD$边上的一点(不与点$A$、点$D$重合),将正方形纸片折叠,使点$B$落在点$P$处,点$C$落在点$G$处,$PG$交$DC$于点$H$,折痕为$EF$,连接$BP$,$BH$.
(1)求证:$∠ APB = ∠ BPH$;
(2)当点$P$在边$AD$上移动时,$△ PDH$的周长是否发生变化?并证明你的结论.

(1)求证:$∠ APB = ∠ BPH$;
(2)当点$P$在边$AD$上移动时,$△ PDH$的周长是否发生变化?并证明你的结论.
答案
26. (1) 证明:如图(1),$\because PE = BE$,$\therefore ∠ EBP = ∠ EPB$. 又 $\because ∠ EPH = ∠ EBC = 90^{\circ}$,$\therefore ∠ EPH - ∠ EPB = ∠ EBC - ∠ EBP$,即 $∠ PBC = ∠ BPH$. 又 $\because AD // BC$,$\therefore ∠ APB = ∠ PBC$. $\therefore ∠ APB = ∠ BPH$.
(2) $△ PDH$ 的周长不变为定值 8. 证明:如图(2),过点 B 作 $BQ ⊥ PH$,垂足为 Q. 由(1)知 $∠ APB = ∠ BPH$,又 $\because ∠ A = ∠ BQP = 90^{\circ}$,$BP = BP$,$\therefore △ ABP ≌ △ QBP$. $\therefore AP = QP$,$AB = BQ$. 又 $\because AB = BC$,$\therefore BC = BQ$. 又 $\because ∠ C = ∠ BQH = 90^{\circ}$,$BH = BH$,$\therefore △ BCH ≌ △ BQH$. $\therefore CH = QH$. $\therefore △ PHD$ 的周长为:$PD + DH + PH = AP + PD + DH + HC = AD + CD = 8$.
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