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7. 如图,在四边形ABCD中,$∠B=∠C=90^{\circ },AB＞CD,AD=AB+CD.$
(1)利用尺规作$∠ADC$的平分线DE,交BC于点E,连接AE(不写作法,保留作图痕迹);
(2)在(1)的条件下,求证:$AE⊥DE.$
(1)利用尺规作$∠ADC$的平分线DE,交BC于点E,连接AE(不写作法,保留作图痕迹);
(2)在(1)的条件下,求证:$AE⊥DE.$
答案
7.
(1) 如图所示
(2) 如图，在 $DA$ 上截取 $GD = CD$，连接 $GE$. $\because DE$ 是$\angle ADC$的平分线，$\therefore \angle GDE = \angle CDE$. 在$\triangle GDE$和$\triangle CDE$中，$\begin{cases} GD = CD, \\ \angle GDE = \angle CDE, \\ DE = DE, \end{cases}$ $\therefore \triangle GDE \cong \triangle CDE$ (SAS)，$\therefore \angle DGE = \angle C = 90°$，$\angle DEG = \angle DEC = \frac{1}{2} \angle CEG$，$\therefore \angle AGE = 180° - \angle DGE = 90°$，$\therefore \angle AGE = \angle B = 90°$，$\therefore \triangle AGE$和$\triangle ABE$均是直角三角形. $\because AD = AG + GD = AB + CD$，$GD = CD$，$\therefore AG = AB$. 在$Rt \triangle AEG$和$Rt \triangle AEB$中，$\begin{cases} AE = AE, \\ AG = AB, \end{cases}$ $\therefore Rt \triangle AEG \cong Rt \triangle AEB$ (HL)，$\therefore \angle AEG = \angle AEB = \frac{1}{2} \angle BEG$，$\therefore \angle AED = \angle DEG + \angle AEG = \frac{1}{2} (\angle CEG + \angle BEG) = \frac{1}{2} × 180° = 90°$，$\therefore AE \perp DE$
8. (新考法·探究题)已知CD是经过$∠BCA$的顶点C的一条直线,$CA=CB$,E,F分别是直线CD上的两点,且$∠BEC=∠CFA=∠α$.
(1)如图①,若直线CD经过$∠BCA$的内部,且点E,F在射线CD上,$0^{\circ }＜∠BCA＜180^{\circ },∠α+∠BCA=180^{\circ }$,请写出线段EF,BE,AF之间的数量关系,并证明你的结论;
(2)如图②,若直线CD不经过$∠BCA$的内部,$∠α=∠BCA$,请写出线段EF,BE,AF之间的数量关系,并证明你的结论.
(1)如图①,若直线CD经过$∠BCA$的内部,且点E,F在射线CD上,$0^{\circ }＜∠BCA＜180^{\circ },∠α+∠BCA=180^{\circ }$,请写出线段EF,BE,AF之间的数量关系,并证明你的结论;
(2)如图②,若直线CD不经过$∠BCA$的内部,$∠α=∠BCA$,请写出线段EF,BE,AF之间的数量关系,并证明你的结论.
答案
8.
(1) $EF = BE - AF$ $\because \angle \alpha + \angle BCA = 180°$，$\therefore \angle \alpha + \angle BCE + \angle ACF = 180°$. $\because \triangle ACF$的内角和为$180°$，$\therefore \angle \alpha + \angle ACF + \angle CAF = 180°$，$\therefore \angle BCE = \angle CAF$. 在$\triangle BCE$和$\triangle CAF$中，$\begin{cases} \angle BEC = \angle CFA, \\ \angle BCE = \angle CAF, \\ CB = AC, \end{cases}$ $\therefore \triangle BCE \cong \triangle CAF$ (AAS)，$\therefore BE = CF$，$CE = AF$. $\because CF = CE + EF$，$\therefore EF = CF - CE = BE - AF$
(2) $EF = BE + AF$ 根据题意，得$\angle BEC = \angle CFA = \angle \alpha = \angle BCA$. 又$\because \angle EBC + \angle BCE + \angle BEC = 180°$，$\angle BCE + \angle FCA + \angle BCA = 180°$，$\therefore \angle EBC = \angle FCA$. 在$\triangle BEC$和$\triangle CFA$中，$\begin{cases} \angle BEC = \angle CFA, \\ \angle EBC = \angle FCA, \\ CB = AC, \end{cases}$ $\therefore \triangle BEC \cong \triangle CFA$ (AAS)，$\therefore CE = AF$，$BE = CF$，$\therefore EF = CF + CE = BE + AF$
(1) $EF = BE - AF$ $\because \angle \alpha + \angle BCA = 180°$，$\therefore \angle \alpha + \angle BCE + \angle ACF = 180°$. $\because \triangle ACF$的内角和为$180°$，$\therefore \angle \alpha + \angle ACF + \angle CAF = 180°$，$\therefore \angle BCE = \angle CAF$. 在$\triangle BCE$和$\triangle CAF$中，$\begin{cases} \angle BEC = \angle CFA, \\ \angle BCE = \angle CAF, \\ CB = AC, \end{cases}$ $\therefore \triangle BCE \cong \triangle CAF$ (AAS)，$\therefore BE = CF$，$CE = AF$. $\because CF = CE + EF$，$\therefore EF = CF - CE = BE - AF$
(2) $EF = BE + AF$ 根据题意，得$\angle BEC = \angle CFA = \angle \alpha = \angle BCA$. 又$\because \angle EBC + \angle BCE + \angle BEC = 180°$，$\angle BCE + \angle FCA + \angle BCA = 180°$，$\therefore \angle EBC = \angle FCA$. 在$\triangle BEC$和$\triangle CFA$中，$\begin{cases} \angle BEC = \angle CFA, \\ \angle EBC = \angle FCA, \\ CB = AC, \end{cases}$ $\therefore \triangle BEC \cong \triangle CFA$ (AAS)，$\therefore CE = AF$，$BE = CF$，$\therefore EF = CF + CE = BE + AF$
