旋轉(zhuǎn)和軸對(duì)稱等圖形變換是數(shù)學(xué)中考中永恒的話題. 本文選取一道有關(guān)軸對(duì)稱問題的中考題,探究其解題策略.
原題再現(xiàn)
例 (2023·遼寧·沈陽)如圖[1],在[?ABCD]紙片中,[AB=10],[AD=6],[∠DAB=60°],點(diǎn)[E]為[BC]邊上的一點(diǎn)(點(diǎn)[E]不與點(diǎn)[C]重合),連接[AE],將[?ABCD]紙片沿[AE]所在直線折疊,點(diǎn)[C],[D]的對(duì)應(yīng)點(diǎn)分別為[C'],[D'],射線[C'E]與射線[AD]交于點(diǎn)[F].
(1)求證:[AF=EF];
(2)如圖[2],當(dāng)[EF⊥AF]時(shí),[DF]的長為 ;
(3)如圖[3],當(dāng)[CE=2]時(shí),過點(diǎn)[F]作[FM⊥AE],垂足為點(diǎn)[M],延長[FM]交[C'D']于點(diǎn)[N],連接[AN],[EN],求[△ANE]的面積.
破解思路
第(1)問證明兩條線段相等有兩種基本策略:一是利用等角對(duì)等邊,二是證明兩個(gè)三角形全等. 在此問中,AF與EF在△AEF中,所以通過證明∠FAE = ∠FEA來解答.
思路1:基于∠FAE = ∠D′AE,結(jié)合EC′ [?] AD′進(jìn)行證明.
解法1:∵四邊形AD′C′E是由四邊形ADCE沿直線AE折疊得到的,
∴∠FAE = ∠D′AE.
∵四邊形ABCD是平行四邊形,
∴CE [?] AD,∴EC′ [?] AD′,
∴∠D′AE = ∠FEA,∴∠FAE = ∠FEA,∴AF = EF.
思路2:基于∠AEC = ∠AEC′,結(jié)合EC′ [?] AD′進(jìn)行證明.
解法2:∵四邊形AD′C′E是由四邊形ADCE沿直線AE折疊得到的,
∴∠AEC = ∠AEC′,∴∠CEF + ∠AEF = ∠BEC′ + ∠AEB.
∵∠CEF = ∠BEC′,∴∠AEF = ∠AEB.
∵四邊形ABCD是平行四邊形,∴BC [?] AD,
∴∠FAE = ∠AEB,∴∠FAE = ∠AEF,∴AF = EF.
第(2)問求線段的長度,有兩種解題策略:一是在直角三角形中通過三角函數(shù)計(jì)算;二是通過第(1)問的結(jié)論,構(gòu)造直角三角形,利用特殊角計(jì)算.
思路1:在(1)的結(jié)論下可知△AEF是等腰直角三角形,結(jié)合∠C = ∠DAB = 60°,用DF分別表示出EF和AF,建立方程求出DF.
解法1:如圖4,設(shè)DF = x,EF與CD的交點(diǎn)為G.
∵四邊形ABCD是平行四邊形,
∴AB [?] CD,AD [?] BC,∠C = ∠DAB = 60°,
∴∠FDG = ∠DAB = 60°.
∵EF ⊥ AF,∴∠F = 90°,∴∠CEF = 90°,
思路2:同樣在(1)中已經(jīng)證明了AF = EF,由(2)中給出的EF ⊥ AF,我們不難得出實(shí)際此時(shí)EF是平行四邊形ABCD的高.所以只要求出平行四邊形ABCD中AD邊上的高即可.
解法2:如圖5,過點(diǎn)B作BH [⊥] AD于點(diǎn)H,
∵四邊形ABCD是平行四邊形,∴AD [?] BC.
∵HB ⊥ AD,EF ⊥ AF,∴∠F = ∠AHB = 90°,
∴EF [?] HB,∴四邊形EFHB是矩形,
第(3)問求△ANE的面積,有兩種解題策略:一是利用三角形面積公式;二是將三角形面積比轉(zhuǎn)化為線段比.
思路1:如圖6,由線段CD與C′D′關(guān)于直線AE對(duì)稱,可知分別延長DC與D′C′,其交點(diǎn)K在直線AE上.由題意知EC′ [?] AD′,F(xiàn)N ⊥ AE,得到M,E是線段AK的三等分點(diǎn).過點(diǎn)A作AH ⊥ KD′于點(diǎn)H,可求出△AKH的邊長,再通過△KMN ∽ △KHA,求得MN的長.
解法1:∵AF = FE,F(xiàn)M ⊥ AE,∴AM = ME.
如圖6,分別延長DC與D′C′交于點(diǎn)K.
∵四邊形AD′C′E與四邊形ADCE關(guān)于直線AE對(duì)稱,
∴點(diǎn)K在直線AE上,EC′ = EC = 2,AD′ = AD = 6,C′D′ = CD = 10,∠AD′C′ = ∠ADC = 120°.
∵四邊形ABCD是平行四邊形,
∴KD′ = 15,AM = ME = EK.
過點(diǎn)A作AH ⊥ KD′于點(diǎn)H,
思路2:如圖7,把△AFE沿直線AE折疊,點(diǎn)F的對(duì)稱點(diǎn)F′為AD′與CB的延長線的交點(diǎn).可求出△AF′B的邊長,過點(diǎn)F′作F′Q ⊥ AB于點(diǎn)Q,則可求出△ABF′的面積,再求出△ABE的面積,最后通過F′N與MN的比求出△AMN的面積.
解法2:如圖7,分別延長CB,AD′,延長線交于點(diǎn)F′,連接NF'.
∵四邊形AD′C′E與四邊形ADCE關(guān)于直線AE對(duì)稱,
∴∠F′AE = ∠FAE.
由第(1)問解法2可知∠F′EA = ∠FEA,
∴△AF′E ≌ △AFE,∴AF' = AF = FE = EF′,
∴四邊形AFEF′是菱形,∴F′在FN的延長線上.
設(shè)D′F′ = 2x,則F′A = 6 + 2x,F(xiàn)′B = F'E - BE = AF' - BE = 2 + 2x.
過點(diǎn)F′作F′Q ⊥ AB于點(diǎn)Q.
∵AD [?] EF′,∴∠ABF′ = ∠DAB = 60°,
∴AQ = AB - BQ = 9 - x.
在Rt△AQF′中,由勾股定理得AQ2 + F′Q2 = F′A2,
∵四邊形AFEF′是菱形,∴FC′ [?] D′F′,
總結(jié)提升
1.對(duì)于軸對(duì)稱(翻折)相關(guān)題目,要挖掘圖形中軸對(duì)稱的性質(zhì),結(jié)合其他條件解決問題,如例題中第(1)問就是利用成軸對(duì)稱的兩個(gè)圖形對(duì)應(yīng)角相等,進(jìn)而推導(dǎo)出結(jié)論.
2.根據(jù)結(jié)論的需要構(gòu)造軸對(duì)稱(或補(bǔ)全)圖形,如第(3)問中的解法1和解法2,再綜合幾何圖形的其他性質(zhì)完成解題.
軸對(duì)稱(翻折)隱含了相應(yīng)的角相等、線段相等以及相關(guān)的圖形性質(zhì),同學(xué)們在解決此類題目時(shí)要不斷體會(huì)、反思和總結(jié).