《2025年名校课堂九年级数学上册北师大版》

4. 如图,四边形ABCD是正方形,E是边BC上任意一点,∠AEF= 90°,且EF交正方形外角平分线CF于点F.求证:AE= EF.
证明: 在 AB 上截取 BM = BE, 连接 ME. ∵∠B = 90°, ∴∠BME = ∠BEM = 45°. ∵CF 平分 ∠DCG, ∴∠FCG = 45°. ∴∠AME = ∠ECF = 135°. ∵∠AEF = ∠B = 90°, ∴∠AEB + ∠CEF = ∠AEB + ∠MAE = 90°. ∴∠CEF = ∠MAE. ∵AB = BC, BM = BE, ∴AM = EC. ∴△AME ≌ △ECF (ASA). ∴AE = EF.

答案： 证明: 在 $ AB $ 上截取 $ BM = BE $, 连接 $ ME $. $ \because \angle B = 90^{\circ} $, $ \therefore \angle BME = \angle BEM = 45^{\circ} $. $ \because CF $ 平分 $ \angle DCG $, $ \therefore \angle FCG = 45^{\circ} $. $ \therefore \angle AME = \angle ECF = 135^{\circ} $. $ \because \angle AEF = \angle B = 90^{\circ} $, $ \therefore \angle AEB + \angle CEF = \angle AEB + \angle MAE = 90^{\circ} $. $ \therefore \angle CEF = \angle MAE $. $ \because AB = BC $, $ BM = BE $, $ \therefore AM = EC $. $ \therefore \triangle AME \cong \triangle ECF ( ASA ) $. $ \therefore AE = EF $.

5. (本专题T4变式)如图所示,四边形ABCD是正方形,E是边BC上任意一点,∠AEF= 90°,且AE= EF.求证:CF是正方形ABCD外角的平分线.
证明: 在AB上取一点G, 使得AG=EC, 连接GE. ∵∠AEF = 90°, ∴∠AEB + ∠CEF = 90°. ∵ 四边形ABCD为正方形, ∴ AB = BC, ∠B = ∠BCD = 90°. ∴∠GAE + ∠AEB = 90°. ∴∠GAE = ∠CEF. 又∵AE = EF, ∴△AEG ≌ △EFC (SAS). ∴∠AGE = ∠ECF. ∵ AB = BC, AG = EC, ∴ BG = BE. ∴∠BGE = 45°. ∴∠ECF = ∠AGE = 135°. ∴∠DCF = ∠FCH = 45°. ∴ CF是正方形ABCD外角的平分线.

答案： 证明: 在 $ AB $ 上取一点 $ G $, 使得 $ AG = EC $, 连接 $ GE $. $ \because \angle AEF = 90^{\circ} $, $ \therefore \angle AEB + \angle CEF = 90^{\circ} $. $ \because $ 四边形 $ ABCD $ 为正方形, $ \therefore AB = BC $, $ \angle B = \angle BCD = 90^{\circ} $. $ \therefore \angle GAE + \angle AEB = 90^{\circ} $. $ \therefore \angle GAE = \angle CEF $. 又 $ \because AE = EF $, $ \therefore \triangle AEG \cong \triangle EFC ( SAS ) $. $ \therefore \angle AGE = \angle ECF $. $ \because AB = BC $, $ AG = EC $, $ \therefore BG = BE $. $ \therefore \angle BGE = 45^{\circ} $. $ \therefore \angle ECF = \angle AGE = 135^{\circ} $. $ \therefore \angle DCF = \angle FCH = 45^{\circ} $. $ \therefore CF $ 是正方形 $ ABCD $ 外角的平分线.

6. 如图,正方形ABCD的边长为3,E,F分别是边AB,BC上的点,且∠EDF= 45°.
(1)求证:EF= AE+CF.
(2)当AE= 1时,求EF的长.$\frac{5}{2}$

答案： 解:
(1) 证明: 延长 $ BC $ 至点 $ H $, 使 $ CH = AE $, 连接 $ DH $. $ \because $ 四边形 $ ABCD $ 是正方形, $ \therefore AD = CD $, $ \angle A = \angle DCH = 90^{\circ} $. $ \therefore \triangle DAE \cong \triangle DCH ( SAS ) $. $ \therefore DE = DH $, $ \angle ADE = \angle CDH $. $ \because \angle ADC = 90^{\circ} $, $ \angle EDF = 45^{\circ} $, $ \therefore \angle ADE + \angle FDC = 45^{\circ} $. $ \therefore \angle FDC + \angle CDH = 45^{\circ} $, 即 $ \angle HDF = 45^{\circ} $. $ \therefore \angle EDF = \angle HDF = 45^{\circ} $. 又 $ \because DF = DF $, $ \therefore \triangle EDF \cong \triangle HDF ( SAS ) $. $ \therefore EF = FH $. $ \because FH = CH + CF = AE + CF $, $ \therefore EF = AE + CF $.
(2) 设 $ EF = x $, 则 $ FH = x $. $ \because $ 正方形 $ ABCD $ 的边长为 $ 3 $, $ \therefore AB = BC = 3 $. $ \because AE = 1 $, $ \therefore BE = 2 $, $ CH = 1 $. $ \therefore CF = x - 1 $. $ \therefore BF = BC - CF = 3 - ( x - 1 ) = 4 - x $. 在 $ \mathrm { Rt } \triangle BEF $ 中, $ \because BE ^ { 2 } + BF ^ { 2 } = EF ^ { 2 } $, $ \therefore 2 ^ { 2 } + ( 4 - x ) ^ { 2 } = x ^ { 2 } $, 解得 $ x = \frac { 5 } { 2 } $. $ \therefore EF = \frac { 5 } { 2 } $.