3. 在四边形ABCD中,点E,F分别是边AB,AD上的点,连接CE,CF并延长,分别交DA,BA的延长线于点H,G,连接AC.
(1) 如图①,若四边形ABCD是正方形,∠ECF=45°,求证:△ACG∽△AHC;
(2) 如图②,若四边形ABCD是菱形,BC=6,∠ECF=∠CAD=60°,AE=$\frac{7}{2}$,求AG的长;
(3) 如图③,若四边形ABCD是矩形,AD=2AB=6,CG=CH,∠GCH=45°,求GH².

(1) $\because$ 四边形 $ABCD$ 是正方形, $\therefore ∠ ACD = ∠ ACB = \frac{1}{2}∠ BCD=45°.\because ∠ ECF=45°,\therefore ∠ ACD = ∠ ACB = ∠ ECF$,
$\therefore ∠ ACF + ∠ ACH = ∠ ACF + ∠ DCG=45°,\therefore ∠ DCG = ∠ ACH$, 同理 $∠ ACG = ∠ BCE.\because AD// BC,CD// AB,\therefore ∠ G = ∠ DCG$, $∠ H = ∠ BCH,\therefore ∠ G = ∠ ACH,∠ H = ∠ ACG,\therefore △ ACG ∽ △ AHC$.
(2) $\because$ 四边形 $ABCD$ 是菱形, $\therefore AD// BC,\therefore ∠ ACB = ∠ CAD = 60°,\therefore ∠ ACD = ∠ ACB = 60°.\because ∠ ECF=60°,\therefore ∠ DCF + ∠ ACF = ∠ ACF + ∠ ACE=60°,\therefore ∠ DCF = ∠ ACE$, 同理 $∠ BCE = ∠ ACF.\because CD// AB,AD// BC,\therefore ∠ DCF = ∠ G$, $∠ BCE = ∠ H,\therefore ∠ ACE = ∠ G,∠ ACF = ∠ H,\therefore △ ACG ∽ △ AHC,\therefore \frac{AC}{AH}=\frac{AG}{AC},\therefore AC^2=AH· AG.\because BC=AB=6,\therefore AC=6$,
$\therefore AH=\frac{36}{AG}.\because BC// AD,\therefore △ EAH ∽ △ EBC,\therefore \frac{AE}{BE}=\frac{AH}{BC}.\because AE=\frac{7}{2},\therefore BE=AB-AE=6-\frac{7}{2}=\frac{5}{2},\therefore \frac{\frac{7}{2}}{\frac{5}{2}}=\frac{\frac{36}{AG}}{6}$, 解得 $AG=\frac{30}{7}$.
(3) 如图, 取 $BC$ 中点 $M$, 过点 $M$ 作 $MN// BG$, 交 $AD$ 于点 $P$, 交 $CG$ 于点 $N$, 连接 $CP.\because MN// BG,\therefore △ CMN ∽ △ CBG,\therefore \frac{CM}{CB}=\frac{CN}{CG}=\frac{MN}{BG}$. 又 $\because$ 点 $M$ 是 $BC$ 中点,
$\therefore \frac{CM}{CB}=\frac{CN}{CG}=\frac{MN}{BG}=\frac{1}{2},\therefore CB=2CM,CG=2CN,BG=2MN.\because CG=CH,\therefore CG=CH=2CN.\because CD// BA,MN// BG,\therefore CD// MN// BG,\therefore \frac{MC}{MB}=\frac{DP}{PA}=1,\therefore DP=PA.\because AD=2AB=6,\therefore AB=CD=3,AD=BC=6,\therefore CM=DP=3$. 又 $\because BC// AD,\therefore$ 四边形 $CDPM$ 是平行四边形, 且 $CD=DP=3,∠ D=90°,\therefore$ 四边形 $CDPM$ 是正方形, $\therefore CP=3\sqrt{2}.\because$ 四边形 $CDPM$ 是正方形, 且 $∠ GCH=45°$, 由(1) 可得 $△ CPN ∽ △ HPC,\therefore \frac{PH}{PC}=\frac{CP}{NP}=\frac{CH}{NC}=2,\therefore PH=2CP=6\sqrt{2},PN=\frac{1}{2}CP=\frac{3\sqrt{2}}{2},\therefore MN=3+\frac{3\sqrt{2}}{2},AH=PH-PA=6\sqrt{2}-3,\therefore BG=2MN=6+3\sqrt{2},\therefore AG=BG-AB=3+3\sqrt{2}$. 在 $\mathrm{Rt}△ AHG$ 中, $GH^2=AG^2+AH^2=(3+3\sqrt{2})^2+(6\sqrt{2}-3)^2=108-18\sqrt{2}$.

4. 综合与探究:如图,∠AOB=90°,点P在∠AOB的平分线上,PA⊥OA于点A.
(1)【操作判断】
如图①,过点P作PC⊥OB于点C,根据题意在图①中画出PC,图中∠APC的度数为度;
(2)【问题探究】
如图②,点M在线段AO上,连接PM,过点P作PN⊥PM交射线OB于点N,求证:OM+ON=2PA;
(3)【拓展延伸】
点M在射线AO上,连接PM,过点P作PN⊥PM交射线OB于点N,射线NM与射线PO相交于点F.若ON=3OM,求$\frac{OP}{OF}$的值.

(1) 如图①, $PC$ 即为所求. $90$ 解析: $\because ∠ AOB=90°,PA⊥ OA,PC⊥ OB,\therefore$ 四边形 $OAPC$ 是矩形, $\therefore ∠ APC=90°$.

(2) 过点 $P$ 作 $PC⊥ OB$ 于点 $C$, 如图②. 由(1) 知四边形 $OAPC$ 是矩形, $\because$ 点 $P$ 在 $∠ AOB$ 的平分线上, $PA⊥ OA,PC⊥ OB,\therefore PA=PC,\therefore$ 矩形 $OAPC$ 是正方形, $\therefore OA=AP=PC=OC,∠ APC=90°.\because PN⊥ PM,\therefore ∠ APM = ∠ CPN=90°-∠ MPC$. 又 $\because ∠ A = ∠ PCN=90°,AP=CP,\therefore △ APM ≌ △ CPN$ (ASA), $\therefore AM=CN,\therefore OM+ON=OM+CN+OC=OM+AM+PA=OA+PA=2PA$.
(3) ①当点 $M$ 在线段 $AO$ 上时, 如图③, 延长 $NM,PA$ 相交于点 $G$.

由(2) 知 $OM+ON=2PA$, 设 $OM=x$, 则 $ON=3x,AO=PA=2x$,
$\therefore AM=AO-OM=OM=x.\because ∠ AOB = ∠ MAG=90°,∠ AMG = ∠ OMN,\therefore △ AMG ≌ △ OMN$ (ASA), $\therefore AG=ON=3x$.
$\because ∠ AOB=90°,PA⊥ OA,\therefore AP// OB,\therefore △ ONF ∽ △ PGF$,
$\therefore \frac{OF}{PF}=\frac{ON}{PG}=\frac{3x}{3x+2x}=\frac{3}{5},\therefore \frac{PF}{OF}=\frac{5}{3},\therefore \frac{OP}{OF}=\frac{5+3}{3}=\frac{8}{3}$.
②当点 $M$ 在 $AO$ 的延长线上时, 如图④, 过点 $P$ 作 $PC⊥ OB$ 于点 $C$, 并延长交 $MN$ 于点 $G$,

由(2) 知四边形 $OAPC$ 是正方形, $\therefore OA=AP=PC=OC$, $∠ APC=90°,PC// AO.\because PN⊥ PM,\therefore ∠ APM = ∠ CPN=90°-∠ MPC$. 又 $\because ∠ A = ∠ PCN=90°,AP=CP,\therefore △ APM ≌ △ CPN$ (ASA), $\therefore AM=CN,\therefore ON-OM=OC+CN-OM=AO+AM-OM=AO+AO=2AO$. 设 $ON=3OM=3x,\therefore AO=x,CN=AM=2x$.
$\because PC// AO,\therefore △ CGN ∽ △ OMN,\therefore \frac{CG}{OM}=\frac{CN}{ON}$, 即 $\frac{CG}{x}=\frac{2x}{3x}$,
$\therefore CG=\frac{2}{3}x.\because PC// AO,\therefore △ OMF ∽ △ PGF,\therefore \frac{OF}{PF}=\frac{OM}{PG}=\frac{x}{x+\frac{2}{3}x}=\frac{3}{5},\therefore \frac{PF}{OF}=\frac{5}{3},\therefore \frac{OP}{OF}=\frac{5-3}{3}=\frac{2}{3}$. 综上, $\frac{OP}{OF}$ 的值为 $\frac{2}{3}$ 或 $\frac{8}{3}$.