14[2024海南中考]正方形ABCD中,点E是边BC上的动点(不与点B,C重合),∠1=∠2,AE=EF,AF交CD于点H,FG⊥BC交BC延长线于点G.
(1)如图(1),求证:△ABE≌△EGF;
(2)如图(2),EM⊥AF于点P,交AD于点M.
①求证:点P在∠ABC的平分线上;
②当$\frac{CH}{DH}=m$时,猜想AP与PH的数量关系,并证明;
③作HN⊥AE于点N,连结MN,HE,当MN//HE时,若AB=6,求BE的值.

14. (1)【证明】$\because$ 四边形 $ABCD$ 是正方形, $\therefore ∠ ABE=90°$.
$\because FG ⊥ BC, \therefore ∠ EGF=90°$.
$\because ∠ 1=∠ 2, AE=EF, \therefore △ ABE ≌ △ EGF(\mathrm{AAS})$.
(2) ①【证明】如图(1), 连结 $BP$.
由(1)得 $△ ABE ≌ △ EGF, \therefore ∠ AEB = ∠ EFG, \therefore ∠ AEB + ∠ GEF = ∠ AEB + ∠ BAE = 90°$, 则 $∠ AEF=90°$.
$\because AE=EF, \therefore △ AEF$ 是等腰直角三角形. $\because EM ⊥ AF, \therefore ∠ APE=90°, \therefore ∠ AEP = ∠ FEP=45°$.
又 $\because ∠ ABE=90°, \therefore A,B,E,P$ 四点共圆, $\therefore ∠ ABP = ∠ AEP=45°. \because ∠ ABE=90°, \therefore ∠ ABP = ∠ CBP=45°, \therefore$ 点 $P$ 在 $∠ ABC$ 的平分线上.
②【解】$AP=(m+1)PH$. 证明如下:
由①得点 $P$ 在 $∠ ABC$ 的平分线上即正方形 $ABCD$ 的对角线上. 如图(1), 连结 $PD$, 则 $B,P,D$ 三点共线.
$\because$ 正方形 $ABCD$ 中, $AB// HD, \therefore △ ABP ∽ △ HDP, \therefore \frac{AP}{HP}=\frac{AB}{HD}. \because \frac{CH}{DH}=m$,
即 $HC=mHD, \therefore DC=DH+HC=(m+1)HD, \therefore \frac{AP}{HP}=\frac{AB}{HD}=\frac{DC}{HD}=m+1, \therefore AP=(m+1)PH$.
③【解】如图(2), 连结 $PD, MH$. 由①得点 $P$ 在 $∠ ABC$ 的平分线上即正方形 $ABCD$ 的对角线上, $\therefore ∠ PDH=45°$. 同①可得 $M,D,H,P$ 四点共圆, 则 $∠ PMH = ∠ PDH=45°. \because ∠ NEM=45°, \therefore ∠ EMH = ∠ NEM=45°, \therefore MH// EN. \because MN// HE, \therefore$ 四边形 $MNEH$ 是平行四边形. 设平行四边形 $MNEH$ 的对角线的交点为 $Q, \therefore ME=2MQ$.
易证 $△ PHQ$ 和 $△ PHM$ 都是等腰直角三角形, $\therefore$ 设 $PM=PH=PQ=a$, 则 $MQ=2a, ME=2MQ=4a$.
$\because PM=PH$, 且易得 $PA=PE, \therefore AH=ME=4a, \therefore AP=3a$, 则 $AE=3\sqrt{2}a, \therefore BE=\sqrt{AE^2-AB^2}=\sqrt{(3\sqrt{2}a)^2-6^2}=\sqrt{18a^2-36}. \because ∠ APM = ∠ ADH, ∠ PAM = ∠ DAH, \therefore △ APM ∽ △ ADH, \therefore \frac{DH}{AD}=\frac{PM}{AP}=\frac{a}{3a}=\frac{1}{3}, \therefore DH=\frac{1}{3}AD=2, \therefore AH=\sqrt{DH^2+AD^2}=\sqrt{40}=2\sqrt{10}. \because AH=4a, \therefore 2\sqrt{10}=4a, \therefore a=\frac{\sqrt{10}}{2}, \therefore BE=\sqrt{18a^2-36}=\sqrt{18×(\frac{\sqrt{10}}{2})^2-36}=\sqrt{9}=3$, 即 $BE$ 的值为 3.

15[2025浙江杭州期中]如图(1),已知在Rt△ABC中,∠ABC=90°,BC=15,AB=20,D是边AC上一动点,连结BD,将△CBD沿BD翻折至△C'BD.
(1)当C'D//BC时.
①求证:CB=CD;
②求折痕BD的长.
(2)如图(2),当直线C'D与BC垂直时,以B为原点,直线BC为x轴,直线AB为y轴建立平面直角坐标系,求此时C'的坐标.

15. (1) ①【证明】$\because C'D// BC, \therefore ∠ C'DB = ∠ DBC$.
由翻折得 $∠ DBC = ∠ DBC', BC=BC', CD=C'D, \therefore ∠ C'DB = ∠ DBC', \therefore BC'=C'D, \therefore BC=CD$.
【解】②设 $AB$ 交 $C'D$ 于 $K$, 如图(1).
$\because ∠ ABC=90°, BC=15, AB=20, \therefore AC=\sqrt{BC^2+AB^2}=25. \because CD=BC=15, \therefore AD=AC-CD=10. \because ∠ ABC=90°, C'D// BC, \therefore ∠ AKD = ∠ ABC=90° = ∠ BKD. \because ∠ A = ∠ A, \therefore △ ADK ∽ △ ACB, \therefore \frac{AD}{AC}=\frac{DK}{BC}=\frac{AK}{AB}$, 即 $\frac{10}{25}=\frac{DK}{15}=\frac{AK}{20}, \therefore DK=6, AK=8, \therefore BK=AB-AK=20-8=12, \therefore BD=\sqrt{BK^2+DK^2}=\sqrt{12^2+6^2}=6\sqrt{5}, \therefore$ 折痕 $BD$ 的长为 $6\sqrt{5}$.
(2)如图(2), 延长 $C'D$ 交 $BC$ 于 $H$, 则 $C'H ⊥ BC, \therefore ∠ DHC = ∠ ABC=90°. \because ∠ DCH = ∠ ACB, \therefore △ DCH ∽ △ ACB, \therefore \frac{CD}{AC}=\frac{DH}{AB}=\frac{CH}{BC}$, 即 $\frac{CD}{25}=\frac{DH}{20}=\frac{CH}{15}, \therefore$ 设 $DH=4x, CD=5x, CH=3x$. 由翻折得 $BC'=BC=15, C'D=CD=5x, \therefore C'H=C'D+DH=9x, BH=BC-CH=15-3x. \because BH^2+C'H^2=BC'^2, \therefore (15-3x)^2+(9x)^2=15^2$, 解得 $x=0$(舍去)或 $x=1, \therefore BH=15-3x=12, C'H=9x=9, \therefore C'$ 的坐标为 $(12,9)$.
思路分析
(2)延长 $C'D$ 交 $BC$ 于 $H$, 证明 $△ DCH ∽ △ ACB$, 可得 $\frac{CD}{25}=\frac{DH}{20}=\frac{CH}{15}$, 故设 $DH=4x, CD=5x, CH=3x$, 根据 $BH^2+C'H^2=BC'^2$, 得 $(15-3x)^2+(9x)^2=15^2$, 解出 $x$ 的值, 即可求出 $BH$ 和 $C'H$ 的长, 从而得到 $C'$ 的坐标.