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1. (2024·青海中考改编)如图,在$Rt△ABC$中,D是AC的中点,$∠BDC= 60^{\circ },AC= 6$,则BC的长是()
A. 3
B. 4
C. 5
D. 6
A. 3
B. 4
C. 5
D. 6
答案
A
2. 如图,在$△ABC$中,$∠C= 90^{\circ },AB= 10,AC= 8,$$BC= 6$,线段DE的两个端点D,E分别在边AC,BC上滑动,且$DE= 4$,若M,N分别是DE,AB的中点,则MN的最小值为()
A. 2
B. 3
C. 3.5
D. 4
A. 2
B. 3
C. 3.5
D. 4
答案
B 解析:如图,连接CM,CN,在△ABC中,∠ACB = 90°,AB = 10,AC = 8,BC = 6.∵DE = 4,M,N分别是DE,AB的中点,∴CN = $\frac{1}{2}$AB = 5,CM = $\frac{1}{2}$DE = 2.∵MN ≥ CN - CM,∴当点C,M,N在同一直线上时，MN取最小值,∴MN的最小值为5 - 2 = 3.故选B.
3. 如图,四边形ABCD中,$∠ABD= ∠ACD= 90^{\circ },$P是边AD的中点.若$BC= 3,AD= 8$,则$△BPC$的周长为______.
答案
11
4. (无锡中考)如图,$△ABC$中,$∠ACB= 90^{\circ },$$AB= 8cm$,D是AB的中点.现将$△BCD$沿BA方向平移1cm,得到$△EFG$,FG交AC于H,则GH的长等于______cm.
答案
3 解析:∵∠ACB = 90°,D是AB的中点,∴CD = AD = DB = 4cm,∴∠A = ∠ACD.由平移可知CD // GF,∴∠ACD = ∠AHG,∴∠A = ∠AHG，AG = GH.由平移性质可得AG = 4 - 1 = 3(cm),∴GH = AG = 3cm.
5. (1)如图①,在$△ABC$中,$∠BAC$为钝角,AF,CE都是这个三角形的高,P为AC的中点,若$∠B= 42^{\circ }$,则$∠EPF$的度数为______.
(2)(青岛中考)如图②,在四边形ABCD中,$∠ABC= ∠ADC= 90^{\circ }$,E为对角线AC的中点,连接BE,ED,BD.若$∠BAD= 58^{\circ }$,则$∠EBD$的度数为______.
(2)(青岛中考)如图②,在四边形ABCD中,$∠ABC= ∠ADC= 90^{\circ }$,E为对角线AC的中点,连接BE,ED,BD.若$∠BAD= 58^{\circ }$,则$∠EBD$的度数为______.
答案
(1)96° 解析:∵CE⊥BA,∠B = 42°,∴∠BCE = 48°.∵AF⊥BC,CE⊥BA,P为AC的中点,∴PF = $\frac{1}{2}$AC = PC,PE = $\frac{1}{2}$AC = PC,∴∠PFC = ∠PCF,∠PEC = ∠PCE,∴∠EPF = 2∠PCF + 2∠PCE = 2∠BCE = 96°.
(2)32° 解析:∵∠ABC = ∠ADC = 90°,E为对角线AC的中点,∴AE = BE = DE,∴∠BAE = ∠ABE,∠DAE = ∠ADE,∴∠BED = ∠BAE + ∠ABE + ∠DAE + ∠ADE = 2∠BAD = 116°.∵DE = BE = $\frac{1}{2}$AC,∴∠EBD = ∠EDB = $\frac{1}{2}$×(180° - 116°) = 32°.
(2)32° 解析:∵∠ABC = ∠ADC = 90°,E为对角线AC的中点,∴AE = BE = DE,∴∠BAE = ∠ABE,∠DAE = ∠ADE,∴∠BED = ∠BAE + ∠ABE + ∠DAE + ∠ADE = 2∠BAD = 116°.∵DE = BE = $\frac{1}{2}$AC,∴∠EBD = ∠EDB = $\frac{1}{2}$×(180° - 116°) = 32°.
6. (1)如图①,P是$∠AOB$内部任意一点,$PM⊥$$OA,PN⊥OB$,垂足分别是M,N,D是OP的中点,连接DM,DN.求证:$∠MDN= 2∠MON.$
(2)如图②,在$△ABC$中,CD,BE分别是AB,AC边上的高,M是边BC的中点.连接DM,ME,试判断$∠DME与∠A$之间的数量关系,并说明理由.
(3)如图③,将图②的锐角$△ABC变为钝角△ABC$,其余条件不变,直接写出$∠DME与∠BAC$之间的数量关系.
(2)如图②,在$△ABC$中,CD,BE分别是AB,AC边上的高,M是边BC的中点.连接DM,ME,试判断$∠DME与∠A$之间的数量关系,并说明理由.
(3)如图③,将图②的锐角$△ABC变为钝角△ABC$,其余条件不变,直接写出$∠DME与∠BAC$之间的数量关系.
答案
(1)∵PM⊥OA,∴∠OMP = 90°.在Rt△OMP中,D是OP的中点,∴DM = $\frac{1}{2}$OP = DO,∴∠DMO = ∠DOM,∴∠MDP = 2∠MOP.同理可知,∠NDP = 2∠NOP,∴∠MDN = ∠MDP + ∠NDP = 2∠MON.
(2)∠DME = 180° - 2∠A.理由如下:∵CD,BE分别是AB,AC边上的高,M是BC的中点,∴DM = $\frac{1}{2}$BC,ME = $\frac{1}{2}$BC,∴DM = ME.在△ABC中,∠ABC + ∠ACB = 180° - ∠A.∵DM = ME = BM = MC,∴∠BMD + ∠CME = (180° - 2∠ABC) + (180° - 2∠ACB) = 360° - 2(∠ABC + ∠ACB) = 360° - 2(180° - ∠A) = 2∠A,∴∠DME = 180° - 2∠A.
(3)∠DME = 2∠BAC - 180°. 解析:在△ABC中,∠ABC + ∠ACB = 180° - ∠BAC.∵DM = ME = BM = MC,∴∠BME + ∠CMD = 2∠ACB + 2∠ABC = 2(180° - ∠BAC) = 360° - 2∠BAC,∴∠DME = 180° - (360° - 2∠BAC) = 2∠BAC - 180°.
(2)∠DME = 180° - 2∠A.理由如下:∵CD,BE分别是AB,AC边上的高,M是BC的中点,∴DM = $\frac{1}{2}$BC,ME = $\frac{1}{2}$BC,∴DM = ME.在△ABC中,∠ABC + ∠ACB = 180° - ∠A.∵DM = ME = BM = MC,∴∠BMD + ∠CME = (180° - 2∠ABC) + (180° - 2∠ACB) = 360° - 2(∠ABC + ∠ACB) = 360° - 2(180° - ∠A) = 2∠A,∴∠DME = 180° - 2∠A.
(3)∠DME = 2∠BAC - 180°. 解析:在△ABC中,∠ABC + ∠ACB = 180° - ∠BAC.∵DM = ME = BM = MC,∴∠BME + ∠CMD = 2∠ACB + 2∠ABC = 2(180° - ∠BAC) = 360° - 2∠BAC,∴∠DME = 180° - (360° - 2∠BAC) = 2∠BAC - 180°.
