《2025年名校课堂九年级数学上册北师大版》

1. (教材P21例1变式)(2023·黄石)如图,在正方形ABCD中,点M,N分别在AB,BC上,且BM= CN,AN与DM相交于点P.
(1)求证:△ABN≌△DAM.
证明: ∵ 四边形 $ ABCD $ 是正方形, $ \therefore AB = AD = BC $, $ \angle DAM = \angle ABN = 90^{\circ} $. $ \because BM = CN $, $ \therefore BC - CN = AB - BM $, 即 $ BN = AM $. 在 $ \triangle ABN $ 和 $ \triangle DAM $ 中, $ \left\{ \begin{array} { l } { AB = DA, } \\ { \angle ABN = \angle DAM, } \\ { BN = AM, } \end{array} \right. $ $ \therefore \triangle ABN \cong \triangle DAM (SAS) $.
(2)求∠APM的度数.
解: 由 (1) 知, $ \triangle ABN \cong \triangle DAM $, $ \therefore \angle BAN = \angle ADM $. $ \because \angle DAM = 90^{\circ} $，$ \therefore \angle ADM + \angle AMP = 90^{\circ} $，$ \therefore \angle BAN + \angle AMP = \angle ADM + \angle AMP = 90^{\circ} $. $ \therefore \angle APM = 180^{\circ} - ( \angle BAN + \angle AMP ) = 90^{\circ} $.
∠APM的度数为90°.

答案： 解:
(1) 证明:
∵ 四边形 $ ABCD $ 是正方形, $ \therefore AB = AD = BC $, $ \angle DAM = \angle ABN = 90^{\circ} $. $ \therefore \angle ADM + \angle AMP = 90^{\circ} $. $ \because BM = CN $, $ \therefore BC - CN = AB - BM $, 即 $ BN = AM $. 在 $ \triangle ABN $ 和 $ \triangle DAM $ 中, $ \left\{ \begin{array} { l } { AB = DA, } \\ { \angle ABN = \angle DAM, } \\ { BN = AM, } \end{array} \right. $ $ \therefore \triangle ABN \cong \triangle DAM ( SAS ) $.
(2) 由
(1) 知, $ \triangle ABN \cong \triangle DAM $, $ \therefore \angle BAN = \angle ADM $. $ \therefore \angle BAN + \angle AMP = \angle ADM + \angle AMP = 90^{\circ} $. $ \therefore \angle APM = 180^{\circ} - ( \angle BAN + \angle AMP ) = 90^{\circ} $.

2. (本专题T1变式)如图,在正方形ABCD中,点E,F分别在边BC,CD上,AE,BF相交于点O,∠AOF= 90°.求证:BE= CF.
证明: ∵ 四边形 $ ABCD $ 是正方形, $ \therefore AB = BC $, $ \angle ABE = \angle BCF = 90^{\circ} $. $ \because \angle AOB = 180^{\circ} - \angle AOF = 90^{\circ} $, $ \therefore \angle BAE + \angle OBA = 90^{\circ} $. 又 $ \because \angle ABE = \angle CBF + \angle OBA = 90^{\circ} $, $ \therefore \angle BAE = \angle CBF $. $ \therefore \triangle ABE \cong \triangle BCF (ASA) $. $ \therefore BE = CF $.

答案： 证明:
∵ 四边形 $ ABCD $ 是正方形, $ \therefore AB = BC $, $ \angle ABE = \angle BCF = 90^{\circ} $. $ \because \angle AOB = 180^{\circ} - \angle AOF = 90^{\circ} $, $ \therefore \angle BAE + \angle OBA = 90^{\circ} $. 又 $ \because \angle ABE = \angle CBF + \angle OBA = 90^{\circ} $, $ \therefore \angle BAE = \angle CBF $. $ \therefore \triangle ABE \cong \triangle BCF ( ASA ) $. $ \therefore BE = CF $.

3. (教材P25习题T4变式)如图,正方形ABCD的对角线AC和BD相交于点O,O又是正方形$A_1B_1C_1O$的一个顶点$,OA_1$交AB于点$E,OC_1$交BC于点F.
(1)求证:△AOE≌△BOF.
证明: 在正方形 $ ABCD $ 中, $ AO = BO $, $ \angle AOB = \angle A _ { 1 } OC _ { 1 } = 90^{\circ} $, $ \angle OAE = \angle OBF = 45^{\circ} $. $ \therefore \angle AOE + \angle EOB = 90^{\circ} $, $ \angle BOF + \angle EOB = 90^{\circ} $. $ \therefore \angle AOE = \angle BOF $. $ \therefore \triangle AOE \cong \triangle BOF $ (ASA).
(2)如果两个正方形的边长都为a,正方形$A_1B_1C_1O$绕点O转动,那么两个正方形重叠部分(四边形OEBF)的面积等于$\frac{1}{4}a^2$.为什么?
理由: $ \because \triangle AOE \cong \triangle BOF $, $ \therefore S _ { 四边形 OEBF } = S _ { \triangle EOB } + S _ { \triangle BOF } = S _ { \triangle EOB } + S _ { \triangle AOE } = S _ { \triangle AOB } = \frac { 1 } { 4 } S _ { 正方形 ABCD } = \frac { 1 } { 4 } a ^ { 2 } $.

答案： 解:
(1) 证明: 在正方形 $ ABCD $ 中, $ AO = BO $, $ \angle AOB = \angle A _ { 1 } OC _ { 1 } = 90^{\circ} $, $ \angle OAE = \angle OBF = 45^{\circ} $. $ \therefore \angle AOE + \angle EOB = 90^{\circ} $, $ \angle BOF + \angle EOB = 90^{\circ} $. $ \therefore \angle AOE = \angle BOF $. $ \therefore \triangle AOE \cong \triangle BOF ( ASA ) $.
(2) 两个正方形重叠部分的面积等于 $ \frac { 1 } { 4 } a ^ { 2 } $. 理由: $ \because \triangle AOE \cong \triangle BOF $, $ \therefore S _ { 四边形 OEBF } = S _ { \triangle EOB } + S _ { \triangle BOF } = S _ { \triangle EOB } + S _ { \triangle AOE } = S _ { \triangle AOB } = \frac { 1 } { 4 } S _ { 正方形 ABCD } = \frac { 1 } { 4 } a ^ { 2 } $.