2018年中考数学真题分类汇编专题复习(七)几何综合题(答案不全) - 图文

类型2 与图形变换有关的几何综合题

(2018宜昌)在矩形ABCD中,AB?12,P是边AB上一点,把PBC沿直线PC折叠,顶点B的对应点是点

G,过点B作BE?CG,垂足为E且在AD上,BE交PC于点F.

(1)如图1,若点E是AD的中点,求证:?AEB≌?DEC; (2) 如图2,①求证: BP?BF;

②当AD?25,且AE?DE时,求cos?PCB的值; ③当BP?9时,求BE EF的值.

图1 图2 图2备用图 23.(1)证明:在矩形ABCD中,?A??D?90,AB?DC, 如图1,又

AE?DE,

图1

?ABE??DCE,

(2)如图2,

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图2

①在矩形ABCD中,?ABC?90,

?BPC沿PC折叠得到?GPC

??PGC??PBC?90,?BPC??GPCBE?CG ?BE//PG, ??GPF??PFB

??BPF??BFP ?BP?BF

②当AD?25时,

?BEC?90

??AEB??CED?90, ?AEB??ABE?90,

??CED??ABE

?A??D?90,

??ABE∽?DEC ?ABDEAE?CD ∴设AE?x,则DE?25?x,

?12x?25?x12, 解得x1?9,x2?16

AE?DE

?AE?9,DE?16, ?CE?20,BE?15,

由折叠得BP?PG,

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?BP?BF?PG, BE//PG, ??ECF∽?GCP ?EFPG?CECG 设BP?BF?PG?y,

?15?y20y?25 ?y?253 则BP?253 在Rt?PBC中,PC?2510BC3,cos?PCB?PC?252510?310103③若BP?9,

解法一:连接GF,(如图3)

?GEF??BAE?90, BF//PG,BF?PG

∴四边形BPGF是平行四边形

BP?BF,

?平行四边形BPGF是菱形 ?BP//GF, ??GFE??ABE, ??GEF∽?EAB ?EFGF?ABBE ?BEEF?ABGF?12?9?108

解法二:如图2,

?FEC??PBC?90,

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?EFC??PFB??BPF, ??EFC∽?BPC ?EFBP?CECB 又

?BEC??A?90,

由AD//BC得?AEB??EBC,

??AEB∽?EBC ?ABCEBE?CB ?AEEFBE?BP ?BEEF?AEBP?12?9?108

解法三:(如图4)过点F作FH?BC,垂足为HS?BPFS?BF四边形PFEGEF?PG?BFBE图4

BFS?BFCEF?BE?BCEFS?? ?BEC12?BC12?9BE?EF12 ?BEEF?12?9?108

(2018邵阳)

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