登录
18. (10 分)(2026·南京期末) 如图, 在 $△ ABC$ 和 $△ ADE$ 中, $AB = AD$, $AC = AE$, $∠ BAD =$ $∠ CAE$, $DE$ 分别交 $BC$, $AC$ 于点 $F$, $G$, 连接 $AF$.
(1) 求证: $∠ C = ∠ E$;
(2) 求证: $AF$ 平分 $∠ BFE$.
(1) 求证: $∠ C = ∠ E$;
(2) 求证: $AF$ 平分 $∠ BFE$.
答案
18. (1)$\because ∠ BAD=∠ CAE$,$\therefore ∠ BAD+∠ DAC=∠ CAE+∠ DAC$,即$∠ BAC=∠ DAE$.$\because AB=AD$,$AC=AE$,$\therefore △ BAC≌ △ DAE(\mathrm{SAS})$,$\therefore ∠ C=∠ E$.
(2)如图,过点A分别作$AM⊥ BC$于点M,$AN⊥ DE$于点N,由(1)得$△ BAC≌ △ DAE$,$\therefore BC=DE$,$S_{△ BAC}=S_{△ DAE}$,$\therefore \frac{1}{2}BC· AM=\frac{1}{2}DE· AN$,$\therefore AM=AN$.又$\because AM⊥ BC$,$AN⊥ DE$,$\therefore AF$平分$∠ BFE$.
(2)如图,过点A分别作$AM⊥ BC$于点M,$AN⊥ DE$于点N,由(1)得$△ BAC≌ △ DAE$,$\therefore BC=DE$,$S_{△ BAC}=S_{△ DAE}$,$\therefore \frac{1}{2}BC· AM=\frac{1}{2}DE· AN$,$\therefore AM=AN$.又$\because AM⊥ BC$,$AN⊥ DE$,$\therefore AF$平分$∠ BFE$.
19. (10 分) 如图, 已知 $∠ ABC = ∠ ADC = 90°$, $M,N$ 分别是 $AC,BD$ 的中点, 连接 $MB,MD$, $MN$.
(1) 求证:$MN ⊥ BD$;
(2) 若 $∠ BAD = 45°$, 判断 $△ MBD$ 的形状, 并说明理由.
(1) 求证:$MN ⊥ BD$;
(2) 若 $∠ BAD = 45°$, 判断 $△ MBD$ 的形状, 并说明理由.
答案
19. (1)$\because ∠ ABC=∠ ADC=90°$,$M$,$N$分别是$AC$,$BD$的中点,$\therefore$在$\mathrm{Rt}△ ABC$中,$BM=\frac{1}{2}AC$,在$\mathrm{Rt}△ ACD$中,$DM=\frac{1}{2}AC$,$\therefore BM=DM$.又$\because N$是$BD$的中点,$\therefore MN⊥ BD$.
(2)$△ MBD$是等腰直角三角形.理由:$\because M$是$AC$的中点,$\therefore AM=\frac{1}{2}AC=BM$,$\therefore ∠ BAM=∠ ABM$,$\therefore ∠ BMC=2∠ BAM$.同理可得$∠ DMC=2∠ DAM$.又$\because ∠ BAD=45°$,$\therefore ∠ BMD=∠ BMC+∠ DMC=2(∠ BAM+∠ DAM)=2∠ BAD=90°$.又$\because BM=DM$,$\therefore △ MBD$是等腰直角三角形.
(2)$△ MBD$是等腰直角三角形.理由:$\because M$是$AC$的中点,$\therefore AM=\frac{1}{2}AC=BM$,$\therefore ∠ BAM=∠ ABM$,$\therefore ∠ BMC=2∠ BAM$.同理可得$∠ DMC=2∠ DAM$.又$\because ∠ BAD=45°$,$\therefore ∠ BMD=∠ BMC+∠ DMC=2(∠ BAM+∠ DAM)=2∠ BAD=90°$.又$\because BM=DM$,$\therefore △ MBD$是等腰直角三角形.
20. (10 分)观察与类比：
(1) 如图①,在$△ ABC$中,$∠ ACB=90°$,点$D$在$△ ABC$外,连接$AD$,作$DE ⊥ AB$于点$E$,交$BC$于点$F$,$AD=AB$,$AE=AC$,连接$AF$,求证：$DF=BC+CF$;
(2) 如图②,$AB=AD$,$AC=AE$,$∠ ACB=∠ AED=90°$,延长$BC$交$DE$于点$F$,写出$DF$,$BC$,$CF$之间的数量关系,并证明你的结论.
(1) 如图①,在$△ ABC$中,$∠ ACB=90°$,点$D$在$△ ABC$外,连接$AD$,作$DE ⊥ AB$于点$E$,交$BC$于点$F$,$AD=AB$,$AE=AC$,连接$AF$,求证：$DF=BC+CF$;
(2) 如图②,$AB=AD$,$AC=AE$,$∠ ACB=∠ AED=90°$,延长$BC$交$DE$于点$F$,写出$DF$,$BC$,$CF$之间的数量关系,并证明你的结论.
答案
20. (1)$\because DE⊥ AB$,$∠ ACB=90°$,$\therefore ∠ AED=∠ AEF=∠ ACB=90°$.在$\mathrm{Rt}△ ACF$与$\mathrm{Rt}△ AEF$中,$\begin{cases} AC=AE,\\ AF=AF, \end{cases}$$\therefore \mathrm{Rt}△ ACF≌ \mathrm{Rt}△ AEF(\mathrm{HL})$,$\therefore CF=EF$.在$\mathrm{Rt}△ ADE$与$\mathrm{Rt}△ ABC$中,$\begin{cases} AD=AB,\\ AE=AC, \end{cases}$$\therefore \mathrm{Rt}△ ADE≌ \mathrm{Rt}△ ABC(\mathrm{HL})$,$\therefore DE=BC$.$\because DF=DE+EF$,$\therefore DF=BC+CF$.
(2)$BC=CF+DF$.证明:连接$AF$,在$\mathrm{Rt}△ ABC$与$\mathrm{Rt}△ ADE$中,$\begin{cases} AB=AD,\\ AC=AE, \end{cases}$$\therefore \mathrm{Rt}△ ABC≌ \mathrm{Rt}△ ADE(\mathrm{HL})$,$\therefore BC=DE$.$\because ∠ ACB=90°$,$\therefore ∠ ACF=90°=∠ AED$.在$\mathrm{Rt}△ ACF$与$\mathrm{Rt}△ AEF$中,$\begin{cases} AC=AE,\\ AF=AF, \end{cases}$$\therefore \mathrm{Rt}△ ACF≌ \mathrm{Rt}△ AEF(\mathrm{HL})$,$\therefore CF=EF$.$\because DE=EF+DF$,$\therefore BC=CF+DF$.
(2)$BC=CF+DF$.证明:连接$AF$,在$\mathrm{Rt}△ ABC$与$\mathrm{Rt}△ ADE$中,$\begin{cases} AB=AD,\\ AC=AE, \end{cases}$$\therefore \mathrm{Rt}△ ABC≌ \mathrm{Rt}△ ADE(\mathrm{HL})$,$\therefore BC=DE$.$\because ∠ ACB=90°$,$\therefore ∠ ACF=90°=∠ AED$.在$\mathrm{Rt}△ ACF$与$\mathrm{Rt}△ AEF$中,$\begin{cases} AC=AE,\\ AF=AF, \end{cases}$$\therefore \mathrm{Rt}△ ACF≌ \mathrm{Rt}△ AEF(\mathrm{HL})$,$\therefore CF=EF$.$\because DE=EF+DF$,$\therefore BC=CF+DF$.
