吕朋
轴对称、平移和旋转是初中阶段常见的三种图形变换,也是中考数学倒数第二题的必考内容. 本文以2019年沈阳市沈河区二模题为例,介绍此类问题的解法.
例 如图1,在矩形纸片 ABCD中,AB = 2,AD = 6,将纸片沿对角线AC对折,使得点D落在点P处.
(1)填空:∠BCA的大小是 .
(2)如图2,将折叠后的纸片沿着AC剪开,把△APC绕点A逆时针旋转α角 (0°≤ α ≤ 90°),得到△AP′C′,点 P,C分别对应点P′,C′,P′A交BC于点 E,P′C′交CD于点F.
①当α = 15°时,求证:AB = BE;
②填空:当点P′落在边BC上时,连接AF,则tan∠DAF的值为 ;
③填空:在②的条件下,将△AP′C′沿着AP′折叠至△AP′C′′处,点C′对应点C′′,AC′′交BC于点G,则线段BG的长度为 .
分析:(1)借助特殊角三角函数值即可求解;(2)根据“一线三直角”及相似三角形的知识解答即可.
解:(1)在Rt△ABC中,∠B = 90°,∴tan∠BCA = = = ,∴∠BCA = 30°.
(2)①当α = 15°时,由(1)得:∠BCA = 30°,则∠P′AC′ = ∠DAC = ∠BCA = 30°,
∴∠P′AC = ∠P′AC′ - ∠CAC′ = 15°,∴∠BAE = ∠BAC - ∠P′AC = 45°,
又∵∠B = 90°,∴△ABE是等腰直角三角形,AB = BE;
②如图3,当点P′落在边BC上时,可证得Rt△ADF ≌Rt△AP′F(HL),
∴∠DAF = ∠P′AF,
借助“一线三直角”可证△CFP′∽△BP′A,
∴tan∠P′AF = = ,
在Rt△ABP′中,BP′ = = 2,
∴CP′ = BC - BP′ = 6 - 2,
∴tan∠P′AF = = = - ,∴tan∠DAF = - .
③方法一:如圖4,过点G作GQ⊥AP′,垂足为点Q,
在Rt△AC′′P′中,∠C′′AP′ = 30°,
设GQ = x,则AQ = x,P′Q = 6 - x,
∵∠GP′Q = ∠AP′B,∠GQP′ = ∠ABP′ = 90°,∴△GP′Q∽△AP′B,
∴ = = ,即: = = ,
∴GP′ = x, = ,解得x = 6 - 6,
∴BG = BP′ - GP′ = 2 - x = 8 - 18;
方法二:如图5,过点G作GQ⊥C′′P′,垂足为点Q,
在Rt△GQC′′中,∠GC′′Q = 60°,
设C′′Q = x,则GQ = x,P′Q = 2 - x,可证得△GP′Q∽△P′AB,
∴ = = ,即: = = ,
∴GP′ = ,∴ = ,解得x = 6 - 4,
∴BG = BP′ - GP′ = 2 - = 8 - 18;
方法三:如图6,过点C′′作C′′Q⊥BC,垂足为点Q,可证得△C′′P′Q∽△P′AB,
∴ = = ,即: = = ,
∴P′Q = 2,C′′Q = 2,
设BG = x,则GQ = 2 - 2 - x,可证得△C′′GQ∽△AGB,
∴ = ,即 = ,
解得x = 8 - 18.
即BG = 8 - 18.
点评:解题过程中运用了多种方法,涉及“A字型”相似、“X字型”相似、勾股定理、“一线三直角”模型、特殊的三角函数值应用以及函数思想,同学们要注意体会每种方法各自的特点.