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1. 如图，在$△ ABC$中，点$D$，$E$分别在边$AC$，$AB$上，$BD$与$CE$交于点$F$. 若$BF = 5\sqrt{3}$，$CF = 2\sqrt{3}$，$∠ ABD = ∠ ACE$，$AD = 2CD$，则线段$EF$的长为
$\frac{5\sqrt{3}}{2}$
.
答案：
1. $\frac{5\sqrt{3}}{2}$ 点拨:设 $DF = x$，$\because ∠BFE = ∠CFD$，$∠ABD = ∠ACE$，$\therefore △EBF ∽ △DCF$，
$\therefore \frac{EF}{DF} = \frac{BF}{CF} = \frac{5\sqrt{3}}{2\sqrt{3}} = \frac{5}{2}$，则 $EF = \frac{5}{2}DF = \frac{5}{2}x$。
$\because AD = 2CD$，$\therefore \frac{CD}{CA} = \frac{1}{3}$。
如答图，过点 $A$ 作 $AG // BD$ 交 $CE$ 的延长线于点 $G$，
$\therefore ∠CDF = ∠CAG$，$∠CFD = ∠CGA$，
$\therefore △CDF ∽ △CAG$，$\therefore \frac{DF}{AG} = \frac{CF}{CG} = \frac{CD}{CA} = \frac{1}{3}$，
$\therefore AG = 3DF = 3x$，$CG = 3CF = 6\sqrt{3}$，
$\therefore EG = CG - CF - EF = 4\sqrt{3} - \frac{5}{2}x$。
$\because AG // BF$，$\therefore ∠AGE = ∠BFE$，$∠GAE = ∠FBE$，
$\therefore △AGE ∽ △BFE$，
$\therefore \frac{AG}{BF} = \frac{EG}{EF}$，即 $\frac{3x}{5\sqrt{3}} = \frac{4\sqrt{3} - \frac{5}{2}x}{\frac{5}{2}x}$，
解得 $x = \sqrt{3}$（负值已舍去），$\therefore EF = \frac{5}{2}x = \frac{5\sqrt{3}}{2}$。
1. $\frac{5\sqrt{3}}{2}$ 点拨:设 $DF = x$，$\because ∠BFE = ∠CFD$，$∠ABD = ∠ACE$，$\therefore △EBF ∽ △DCF$，
$\therefore \frac{EF}{DF} = \frac{BF}{CF} = \frac{5\sqrt{3}}{2\sqrt{3}} = \frac{5}{2}$，则 $EF = \frac{5}{2}DF = \frac{5}{2}x$。
$\because AD = 2CD$，$\therefore \frac{CD}{CA} = \frac{1}{3}$。
如答图，过点 $A$ 作 $AG // BD$ 交 $CE$ 的延长线于点 $G$，
$\therefore ∠CDF = ∠CAG$，$∠CFD = ∠CGA$，
$\therefore △CDF ∽ △CAG$，$\therefore \frac{DF}{AG} = \frac{CF}{CG} = \frac{CD}{CA} = \frac{1}{3}$，
$\therefore AG = 3DF = 3x$，$CG = 3CF = 6\sqrt{3}$，
$\therefore EG = CG - CF - EF = 4\sqrt{3} - \frac{5}{2}x$。
$\because AG // BF$，$\therefore ∠AGE = ∠BFE$，$∠GAE = ∠FBE$，
$\therefore △AGE ∽ △BFE$，
$\therefore \frac{AG}{BF} = \frac{EG}{EF}$，即 $\frac{3x}{5\sqrt{3}} = \frac{4\sqrt{3} - \frac{5}{2}x}{\frac{5}{2}x}$，
解得 $x = \sqrt{3}$（负值已舍去），$\therefore EF = \frac{5}{2}x = \frac{5\sqrt{3}}{2}$。
2. 如图，在$Rt△ ABC$中，$∠ BAC = 90^{\circ}$，$AB = 3$，$AC = 4$，$P$为$BC$上任意一点，连接$PA$，以$PA$，$PC$为邻边作平行四边形$PAQC$，连接$PQ$，则$PQ$的最小值为
$\frac{12}{5}$
.
答案：
2. $\frac{12}{5}$
3. 如图，在菱形$ABCD$中，点$G$在边$CD$上，连接$AG$并延长交$BC$的延长线于点$F$，连接$BD$交$AF$于点$E$，连接$CE$.
(1) 若$BE = BC$，$∠ ABC = 80^{\circ}$，请直接写出$∠ DAE$的度数；
(2) 求证：$EC^{2} = EF · EG$；
(3) 若$AB = 6$，$\frac{CE}{EG} = 3$，求$CF$的长.
(1) 若$BE = BC$，$∠ ABC = 80^{\circ}$，请直接写出$∠ DAE$的度数；
(2) 求证：$EC^{2} = EF · EG$；
(3) 若$AB = 6$，$\frac{CE}{EG} = 3$，求$CF$的长.
答案：
3.
(1) 解: $\because$ 四边形 $ABCD$ 是菱形，$BE = BC$，
$\therefore AB = BC = BE$，$AD // BC$，$∠ABD = \frac{1}{2}∠ABC = 40^{\circ}$，
$\therefore ∠BAD = 180^{\circ} - ∠ABC = 180^{\circ} - 80^{\circ} = 100^{\circ}$，$∠BAE = \frac{180^{\circ} - ∠ABE}{2} = \frac{180^{\circ} - 40^{\circ}}{2} = 70^{\circ}$，
$\therefore ∠DAE = ∠BAD - ∠BAE = 100^{\circ} - 70^{\circ} = 30^{\circ}$。
(2) 证明: $\because$ 四边形 $ABCD$ 是菱形，
$\therefore DA = DC$，$∠ADE = ∠CDE$。
又 $\because DE = DE$，$\therefore △ADE ≌ △CDE(SAS)$，
$\therefore ∠DAF = ∠DCE$。
又 $\because AD // BC$，$\therefore ∠DAF = ∠F$，$\therefore ∠DCE = ∠F$。
又 $\because ∠CEF = ∠GEC$，$\therefore △CEG ∽ △FEC$，
$\therefore \frac{GE}{EC} = \frac{EC}{EF}$，即 $EC^{2} = EF · EG$。
(3) 解: 在菱形 $ABCD$ 中，$AD = AB = 6$，
设 $GE = a$，则 $AE = CE = 3a$，
$\because EC^{2} = EF · EG$，$\therefore EF = 9a$，
$\therefore AG = AE + EG = 3a + a = 4a$，
$FG = FE - EG = 9a - a = 8a$。
又 $\because AD // BC$，$\therefore ∠ADG = ∠DCF$，$∠DAF = ∠F$，
$\therefore △ADG ∽ △FCG$，$\therefore \frac{CF}{AD} = \frac{FG}{AG} = \frac{8a}{4a} = 2$，
$\therefore CF = 2AD = 2 × 6 = 12$。
(1) 解: $\because$ 四边形 $ABCD$ 是菱形，$BE = BC$，
$\therefore AB = BC = BE$，$AD // BC$，$∠ABD = \frac{1}{2}∠ABC = 40^{\circ}$，
$\therefore ∠BAD = 180^{\circ} - ∠ABC = 180^{\circ} - 80^{\circ} = 100^{\circ}$，$∠BAE = \frac{180^{\circ} - ∠ABE}{2} = \frac{180^{\circ} - 40^{\circ}}{2} = 70^{\circ}$，
$\therefore ∠DAE = ∠BAD - ∠BAE = 100^{\circ} - 70^{\circ} = 30^{\circ}$。
(2) 证明: $\because$ 四边形 $ABCD$ 是菱形，
$\therefore DA = DC$，$∠ADE = ∠CDE$。
又 $\because DE = DE$，$\therefore △ADE ≌ △CDE(SAS)$，
$\therefore ∠DAF = ∠DCE$。
又 $\because AD // BC$，$\therefore ∠DAF = ∠F$，$\therefore ∠DCE = ∠F$。
又 $\because ∠CEF = ∠GEC$，$\therefore △CEG ∽ △FEC$，
$\therefore \frac{GE}{EC} = \frac{EC}{EF}$，即 $EC^{2} = EF · EG$。
(3) 解: 在菱形 $ABCD$ 中，$AD = AB = 6$，
设 $GE = a$，则 $AE = CE = 3a$，
$\because EC^{2} = EF · EG$，$\therefore EF = 9a$，
$\therefore AG = AE + EG = 3a + a = 4a$，
$FG = FE - EG = 9a - a = 8a$。
又 $\because AD // BC$，$\therefore ∠ADG = ∠DCF$，$∠DAF = ∠F$，
$\therefore △ADG ∽ △FCG$，$\therefore \frac{CF}{AD} = \frac{FG}{AG} = \frac{8a}{4a} = 2$，
$\therefore CF = 2AD = 2 × 6 = 12$。
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