6. 如图,在$△ ABC$中,$∠ ABC=60°$,$BC=3$,$AC=4$,$D$,$E$分别为边$AC$,$AB$上两个动点,且$AE=CD$,连接$BD$,$CE$,则$BD+CE$的最小值是________.

6. $\sqrt{37}$

7. 如图,在$△ ABC$中,$∠ ABC=30°$,$AB=2$,$BC=3$,在$△ ABC$内部有一点$P$,连接$PA,PB,PC$,则$PA+PB+PC$的最小值为________.

7. $\sqrt{13}$ 解析:如图,将$△ ABP$绕点$B$逆时针旋转$60°$,得到$△ A'BP'$,连接$PP',A'C$,则$∠ ABA' = ∠ PBP' = 60°$,$A'B=AB=2$,$BP=BP'$,$\therefore △ BP'P$为等边三角形,$\therefore BP = PP'$,$∠ A'BC = ∠ ABA' + ∠ ABC = 90°$,$\therefore PA+PB+PC = A'P'+PP'+CP ≥ A'C$,当且仅当$A',P',P,C$四点共线时,$PA+PB+PC$的值最小,为$A'C$的长.$\because ∠ A'BC = 90°$,$A'B=2$,$BC=3$,$\therefore A'C = \sqrt{A'B^2+BC^2} = \sqrt{13}$,$\therefore PA+PB+PC$的最小值为$\sqrt{13}$.

8. 如图,在$△ ABC$中,$AC=2$,$∠ A=60°$,$BD ⊥ AC$交$AC$于点$D$,$P$为线段$BD$上的动点,则$PC+\dfrac{1}{2}PB$的最小值为________.

8. $\sqrt{3}$ 解析:过点$P$作$PH ⊥ AB$于点$H$,如图①所示.
$\because ∠ A = 60°$,$BD ⊥ AC$,$\therefore ∠ ABD = 30°$,$\therefore PH = \dfrac{1}{2}PB$,$\therefore PC+\dfrac{1}{2}PB = PC+PH$,若使$PC+\dfrac{1}{2}PB$的值最小,也就相当于$PC+PH$的值最小,$\therefore$ 当$C,P,H$三点共线时,$PC+PH$的值最小,如图②,$\because AC=2$,$∠ ACH = 90°-∠ A = 90°-60° = 30°$,$\therefore AH = \dfrac{1}{2}AC = 1$,$\therefore CH = \sqrt{AC^2-AH^2} = \sqrt{2^2-1^2} = \sqrt{3}$,$\therefore PC+\dfrac{1}{2}PB$的最小值为$\sqrt{3}$.

9. 如图,已知$△ ABC$是等边三角形,$AB=4$,动点$M$从点$B$出发,沿射线$BC$方向移动,以$AM$为边在右侧作等边$△ AMN$,取$AC$中点$H$,连接$NH$,则$NH$的最小值为________.

9. $\sqrt{3}$ 解析:如图,连接$CN$,$\because △ ABC$,$△ AMN$是等边三角形,$\therefore AB=AC=4$,$AM=AN$,$∠ B = ∠ BAC = ∠ ACB = ∠ MAN = 60°$,$\therefore ∠ BAM + ∠ MAC = ∠ MAC + ∠ CAN = 60°$,$∠ ACP = 120°$,$\therefore ∠ BAM = ∠ CAN$,$\therefore △ ABM ≌ △ ACN (\mathrm{SAS})$,$\therefore ∠ ACN = ∠ B = 60°$,$BM=CN$,$\therefore ∠ ACN = \dfrac{1}{2}∠ ACP$,$\therefore$ 点$N$在$△ ABC$的外角的平分线上运动,由垂线段最短可知当$NH ⊥ CN$时,$NH$最短.
$\because$ 点$H$是$AC$的中点,$\therefore CH=2$.$\because ∠ ACN = 60°$,$∠ HNC = 90°$,$\therefore ∠ CHN = 30°$,$\therefore CN = \dfrac{1}{2}CH = 1$,$NH = \sqrt{CH^2-CN^2} = \sqrt{3}$.

10. （2025·盐城期中）如图，在$△ ABC$中，$∠ ACB=120°$，$AC=BC=4$，点$D$是$BC$的中点，点$P$是$AC$边上的一个动点，连接$PD$，以$PD$为边在$PD$的下方作等边$△ DPQ$，连接$CQ$，则$CQ$的最小值是________.

10. $\sqrt{3}$ 解析:$\because ∠ ACB = 120°$,$AC=BC=4$,点$D$是$BC$的中点,$\therefore BD=CD=2$.$\because$ 以$PD$为边在$PD$的下方作等边$△ DPQ$,$\therefore DP=PQ=DQ$,$∠ PDQ = 60°$.如图,把$△ DCP$绕点$D$顺时针旋转$60°$得$△ DKQ$,连接$CK$,过点$C$作$CF ⊥ KQ$于点$F$,$\therefore ∠ CDK = 60°$,$DC=DK$,$∠ DKQ = ∠ BCA = 120°$,$\therefore △ DCK$为等边三角形,$\therefore ∠ DKC = ∠ DCK = ∠ CDK = 60°$,$DK=CK=CD=2$,$\therefore ∠ CKQ = 120°-60° = 60° = ∠ DCK$,$\therefore CD // KQ$,$\therefore$ 点$Q$在线段$KQ$上运动.$\because ∠ CFK = 90°$,$\therefore ∠ KCF = 30°$,$\therefore KF = \dfrac{1}{2}CK = 1$,在$\mathrm{Rt}△ CFK$中,由勾股定理,得 $CF = \sqrt{2^2-1^2} = \sqrt{3}$,$\therefore$ 当$Q,F$重合时,$CQ$的值最小为$\sqrt{3}$.