6 如图，$△ ABC$和$△ CDE$都是等边三角形，且点$A,C,E$在同一条直线上，$AD$与$BE$，$BC$分别交于点$F,M$，$BE$与$CD$交于点$N$，连接$MN$.有下列结论:① $AM=BN$;② $△ ABF ≌ △ DNF$;③ $∠ FMC + ∠ FNC = 180°$;④ $\frac{1}{MN}=\frac{1}{AC}+\frac{1}{CE}$.其中，正确的是________(填序号).

6. ①③④

7【问题背景】（1）如图①，点B,C,D在同一直线上，∠B=∠ACE=∠D，求证：$△ ABC ∽ △ CDE$.
【问题探究】（2）在（1）的条件下，若C为BD的中点，求证：$AC^2=AB · AE$.
【拓展运用】（3）如图②，在$△ ABC$中，$∠ BAC=90°$，点O是$△ ABC$的内心.若$OA=2\sqrt{2}$，$OB=\sqrt{2}OC$，则BC的长为.

7. （1）$\because ∠ ACD = ∠ ACE + ∠ ECD = ∠ B + ∠ BAC$,$∠ ACE=∠ B,\therefore ∠ BAC=∠ ECD. \because ∠ B=∠ D,\therefore △ ABC ∽$$△ CDE$
（2）$\because △ ABC ∽ △ CDE,\therefore \dfrac{AB}{CD}=\dfrac{AC}{CE}. \because C$ 为 $BD$ 的中点, $\therefore BC = CD. \therefore \dfrac{AB}{BC}=\dfrac{AC}{CE}.$ 又 $\because ∠ B = ∠ ACE$,$\therefore △ ABC ∽ △ ACE. \therefore \dfrac{AB}{AC}=\dfrac{AC}{AE}. \therefore AC^2=AB · AE$
（3）10 【解析】如图,过点 $O$ 作 $EF ⊥ OA$,交 $AB$ 于点 $E$,交 $AC$ 于点 $F$. $\because$ 点 $O$ 是 $△ ABC$ 的内心, $\therefore ∠ EAO = ∠ FAO$.
$\because AO=AO,∠ AOE=∠ AOF,\therefore △ AOE ≌ △ AOF. \therefore AE=$$AF. \therefore △ AEF$ 是等腰直角三角形. $\therefore OE=OF=\dfrac{1}{2}EF=$$OA=2\sqrt{2},AE=AF=\sqrt{OE^2+OA^2}=\sqrt{(2\sqrt{2})^2+(2\sqrt{2})^2}=4.$
$\because ∠ BAC = 90°, \therefore ∠ AEF = ∠ AFE = 45°. \therefore ∠ BEO=$$∠ OFC=135°.$ 易得 $∠ BOC=135°. \because ∠ BOF = ∠ BEO +$$∠ EBO,∠ BOF = ∠ BOC + ∠ COF,\therefore ∠ COF = ∠ OBE.$
$\therefore △ BOE ∽ △ OCF. \therefore \dfrac{OE}{CF}=\dfrac{BE}{OF}=\dfrac{OB}{OC}. \because OB=\sqrt{2}\ OC$,$\therefore \dfrac{OE}{CF}=\dfrac{BE}{OF}=\dfrac{OB}{OC}=\sqrt{2}. \therefore BE=\sqrt{2}\ OF=\sqrt{2}× 2\sqrt{2}=4,CF=$$\dfrac{OE}{\sqrt{2}}=\dfrac{2\sqrt{2}}{\sqrt{2}}=2. \therefore AB=AE+BE=4+4=8,AC=AF+CF=$$4+2=6. \because ∠ BAC=90°,\therefore △ BAC$ 为直角三角形. $\therefore BC=$$\sqrt{AB^2+AC^2}=\sqrt{8^2+6^2}=10.$

8 如图,在$△ ABC$中,$∠ ACB=90°$,$CD⊥ AB$,垂足为$D$,$F$为线段$CD$上一点,连接$AF$并延长至点$E$,连接$CE$,$BE$.当$∠ ACE=∠ AFC$时,请判断$△ AEB$的形状,并说明理由.

8. $△ AEB$ 是直角三角形 理由: $\because ∠ ACB = 90°, \therefore$ 在$\mathrm{Rt}△ ABC$ 中,$∠ CAB + ∠ ABC=90°. \because CD ⊥ AB,\therefore ∠ ADC=$$90°. \therefore$ 在 $\mathrm{Rt}△ ACD$ 中,$∠ CAD + ∠ ACD=90°. \therefore ∠ ABC=$$∠ ACD. \because ∠ CAB = ∠ DAC, \therefore △ ABC ∽ △ ACD. \therefore \dfrac{AB}{AC}=$$\dfrac{AC}{AD}$, 即 $AC^2 = AD · AB. \because ∠ ACE = ∠ AFC, ∠ CAE=$$∠ FAC,\therefore △ AEC ∽ △ ACF. \therefore \dfrac{AC}{AF}=\dfrac{AE}{AC}$,即 $AC^2=AF · AE.$
$\therefore AF · AE = AD · AB. \therefore \dfrac{AF}{AB}=\dfrac{AD}{AE}. \because ∠ FAD = ∠ BAE$,$\therefore △ AFD ∽ △ ABE. \therefore ∠ ADF = ∠ AEB = 90°. \therefore △ AEB$ 是直角三角形.