一、填空题
1. 如图,点 C 在∠AOB 的内部,∠OCA=∠OCB,∠OCA 与∠AOB 互补.若 AC=1.5,BC=2,则 OC 的长为.

1. $\sqrt{3}$

2. 如图,在$△ ABC$中,点$E,F$分别在边$AB,AC$上,$∠ 1=∠ 2$.若$BC=4,AF=2,CF=3$,则$EF$的长为.

2. $\frac{8}{5}$

3. 如图,在$△ ABC$中,$D$,$E$为边$AB$的三等分点,$EF// DG// AC$,$H$为$AF$与$DG$的交点.若$AC=6$,则$DH$的长为.

3. 1

4. 如图，P为$□ ABCD$边BC上一点，E，F分别为PA，PD上的点，且$PA=3PE$，$PD=3PF$，$△ PEF$，$△ PDC$，$△ PAB$的面积分别记为$S$，$S_1$，$S_2$。若$S=2$，则$S_1+S_2$的值为。

4. 18

二、解答题
5. 如图，$△ ABC$ 和 $△ CDE$ 都是等边三角形，$B,C,E$ 三点在同一条直线上，连接 $BD,AD,BD$ 交 $AC$ 于点 $F$.
(1) 若 $AD^2 = DF · BD$，求证：$AD = BF$.
(2) 若 $∠ BAD = 90°, BE = 6$. 求：
① $\tan ∠ DBE$ 的值；
② $DF$ 的长.

5. (1) $\because AD^2 = DF · BD, \therefore \frac{AD}{BD} = \frac{DF}{DA}.$
$\because ∠ ADF = ∠ BDA, \therefore △ ADF ∽ △ BDA. \therefore ∠ FAD = ∠ ABD. \because △ ABC, △ CDE$ 都是等边三角形, $\therefore AB = AC, ∠ BAC = ∠ ABC = ∠ ACB = ∠ DCE = 60°.$
$\therefore ∠ ACD = 60°. \therefore ∠ ACD = ∠ BAF. \therefore △ ADC ≌ △ BFA. \therefore AD = BF$
(2) ① 过点 $D$ 作 $DG ⊥ BE$ 于点 $G. \because ∠ BAD = 90°, ∠ BAC = 60°, \therefore ∠ DAC = 30°.$
$\because ∠ ACD = 60°, \therefore ∠ ADC = 90°. \therefore DC = \frac{1}{2}AC. \therefore$ 易得 $CE = \frac{1}{2}BC. \because BE = 6, \therefore CE = 2, BC = 4. \therefore$ 易得 $CG = EG = 1, BG = 5, DG = \sqrt{3}. \therefore \tan ∠ DBE = \frac{DG}{BG} = \frac{\sqrt{3}}{5}$
② 在 $\mathrm{Rt}△ BDG$ 中, $\because ∠ BGD = 90°, DG = \sqrt{3}, BG = 5,$
$\therefore BD = \sqrt{DG^2 + BG^2} = 2\sqrt{7}. \because ∠ ABC = ∠ DCE = 60°,$
$\therefore CD // AB. \therefore$ 易得 $△ CDF ∽ △ ABF. \therefore \frac{DF}{BF} = \frac{CD}{AB} = \frac{CE}{BC} = \frac{1}{2}. \therefore \frac{DF}{BD} = \frac{1}{3}. \therefore DF = \frac{2\sqrt{7}}{3}$