1. 如图，$∠ AOB=60°$，点P在边OA上，$OP=12$，M，N两点在边OB上，$PM=PN$。若$MN=2$，则OM的长为（）

A.3
B.4
C.5
D.6

1. C

2. 某停车场采用先进的车辆识别系统，车辆进出时被系统自动识别后栏杆抬起.如图,停车场入口的栏杆AO的长为4 m,栏杆AO从水平位置绕点O顺时针旋转到A'O的位置,在旋转过程中,当栏杆的旋转角∠AOA'为30°时,栏杆端点A升高了 m.

2. 2

3. 已知CD是△ABC的高，∠BAC=2∠BCD，P是直线BC上一点.
（1）当点P在CB的延长线上，且∠APC=60°时，如图①，求证：PB+PC=PA；
（2）当点P在边BC上，且∠APC=60°时，如图②；当点P在边BC上，且∠APC=120°时，如图③，请直接写出线段PB，PC，PA之间的数量关系，不需要证明.

3. (1) 如图①,过点A作$AH ⊥ BC$,垂足为H.因为CD是△ABC的高,所以$∠AHB = ∠AHC = ∠BDC = 90°$,即$∠BAH + ∠ABC = 90°$,$∠BCD + ∠ABC = 90°$.所以$∠BAH = ∠BCD$.因为$∠BAC = 2∠BCD$,所以$∠BAC = 2∠BAH$,即$∠BAH = ∠CAH$.又$AH = AH$,所以$△ABH ≌ △ACH$（ASA）.所以$BH = CH$.因为$∠APC = 60°$,所以$∠PAH = 90° - ∠APC = 30°$.所以$PA = 2PH$.因为$PB = PH - BH$,$PC = PH + HC$,所以$PB + PC = PH - BH + PH + CH = 2PH = PA$.
(2) 题图②: $PC - PB = PA$. 题图③: $PB - PC = PA$. 解析:当点P在边BC上,且$∠APC = 60°$时,如图②,过点A作$AH ⊥ BC$,垂足为H.因为CD是△ABC的高,所以$∠AHB = ∠AHC = ∠BDC = 90°$,即$∠BAH + ∠ABC = 90°$,$∠BCD + ∠ABC = 90°$.所以$∠BAH = ∠BCD$.因为$∠BAC = 2∠BCD$,所以$∠BAC = 2∠BAH$,即$∠BAH = ∠CAH$.又$AH = AH$,所以$△ABH ≌ △ACH$（ASA）.所以$BH = CH$.因为$∠APC = 60°$,所以$∠PAH = 90° - ∠APC = 30°$,即$PA = 2PH$.因为$PB = BH - PH$,$PC = PH + HC$,所以$PC - PB = PH + HC - BH + PH = 2PH = PA$,即$PC - PB = PA$.当点P在边BC上,且$∠APC = 120°$时,如图③,过点A作$AH ⊥ BC$,垂足为H.因为CD是△ABC的高,所以$∠AHB = ∠AHC = ∠BDC = 90°$,即$∠BAH + ∠ABC = 90°$,$∠BCD + ∠ABC = 90°$.所以$∠BAH = ∠BCD$.因为$∠BAC = 2∠BCD$,所以$∠BAC = 2∠BAH$,即$∠BAH = ∠CAH$.又$AH = AH$,所以$△ABH ≌ △ACH$(ASA).所以$BH = CH$.因为$∠APC = 120°$,所以$∠APB = 60°$,即$∠HAP = 30°$.所以$PA = 2PH$.因为$PB = BH + PH$,$PC = HC - PH$,所以$PB - PC = BH + PH - HC + PH = 2PH = PA$,即$PB - PC = PA$.

4. 如图,在$△ ABC$中,AD是边BC上的中线,$AB = 10$,$AD = 8$,$∠ BAD=30°$,求$△ ABC$的面积.

4. 延长AD至点E,使$DE = DA$,连接BE,则$AE = 2AD$.因为AD是边BC上的中线,所以$BD = CD$.又$∠BDE = ∠CDA$,所以$△BDE ≌ △CDA$(SAS).所以$S_{△BDE} = S_{△CDA}$,即$S_{△ABC} = S_{△ABE}$.过点B作$BH ⊥ AE$于点H.又$∠BAD = 30°$,$AB = 10$,所以$BH = \frac{1}{2}AB = 5$.又$AD = 8$,所以$AE = 16$,即$S_{△ABE} = \frac{1}{2}AE · BH = 40$.所以$△ABC$的面积为40.

5. 如图,在四边形ABCD中,∠A=90°,∠B=30°,∠C=120°,CD=3,BC=7,则AD的长为 ()

A.2.5
B.2
C.3
D.3.5

5. B 解析:延长BC,AD交于点E.因为$∠A = 90°$,$∠B = 30°$,所以$BE = 2AE$,$∠E = 60°$.因为$∠BCD = 120°$,所以$∠ADC = 360° - ∠A - ∠B - ∠BCD = 120°$,$∠DCE = 180° - ∠BCD = 60°$,即$∠CDE = 180° - ∠ADC = 60°$.所以$△CDE$是等边三角形,即$CE = DE = CD = 3$.又$BC = 7$,所以$BE = 10$,即$AE = 5$.所以$AD = AE - DE = 2$.

6. 如图，在$△ ABC$中，$AB = AC$，$D$，$E$是$△ ABC$内部两点，$AD$平分$∠ BAC$，$∠ EBC = ∠ E = 60°$。若$BE = 6\ \mathrm{cm}$，$DE = 2\ \mathrm{cm}$，则$BC$的长为（）

A.$4\ \mathrm{cm}$
B.$6\ \mathrm{cm}$
C.$8\ \mathrm{cm}$
D.$12\ \mathrm{cm}$

6. C 解析:分别延长AD,ED,交BC于F,G两点.因为$AB = AC$,$AD$平分$∠BAC$,所以$AF ⊥ BC$,$BC = 2BF$.又$∠EBC = ∠E = 60°$,所以$∠BGE = 180° - ∠EBC - ∠E = 60°$,即$∠E = ∠EBC = ∠BGE$.所以$△BEG$是等边三角形,即$EG = BE = BG$.又$BE = 6\ \mathrm{cm}$,$DE = 2\ \mathrm{cm}$,所以$EG = BG = 6\ \mathrm{cm}$,即$DG = EG - DE = 4\ \mathrm{cm}$.又$∠FDG + ∠BGE = 90°$,所以$∠FDG = 30°$,即$FG = \frac{1}{2}DG = 2\ \mathrm{cm}$.所以$BF = BG - FG = 4\ \mathrm{cm}$,即$BC = 8\ \mathrm{cm}$.