1. ★★★ 如图,在$△ ABC$中,$AB ⊥ AC$,$AD ⊥ BC$,$BE$平分$∠ ABC$,交$AD$于点$E$,$EF // AC$,下列结论一定成立的是 ()

A.$∠ ABE = ∠ DFE$
B.$AE = ED$
C.$AD = DC$
D.$AB = BF$

1. D

2. 如图，AD是$△ ABC$的角平分线，$DE ⊥ AC$，垂足为$E$，$BF // AC$交$ED$的延长线于点$F$，若$BC$恰好平分$∠ ABF$，$AE=2BF$。给出下列四个结论：①$DE=DF$；②$DB=DC$；③$AD ⊥ BC$；④$AC=3BF$。其中正确的结论为（）

A.①②③
B.①②④
C.②③④
D.①②③④

2. D

3. （2026·六安月考）如图，动点C与线段AB构成$△ ABC$，其边长满足$AB=9$，$CA=2a+2$，$CB=2a-3$.点D在$∠ ACB$的平分线上，且$∠ ADC=90°$，则$a$的取值范围是________，$△ ABD$的面积的最大值为________.

3. $a＞\dfrac{5}{2}$，$\dfrac{45}{4}$

4. 如图,在直角三角形ABC中,∠ACB=90°,AC=BC,CE⊥AD,垂足为E,BF//AC交CE的延长线于点F.
(1)求证:△ACD≅△CBF;
(2)若D是BC的中点,求证:AC=2BF.
＞＞ 对点专练 P32

4. (1) $\because BF // AC, ∠ ACB = 90°, \therefore ∠ CBF = ∠ ACB = 90°$,
$\therefore ∠ BCF + ∠ F = 90°$.在直角三角形 $CDE$ 中,$CE ⊥ AD,∠ BCF + ∠ ADC = 90°$, $\therefore ∠ F = ∠ ADC$. 在 $△ ACD$ 和 $△ CBF$ 中,
$\begin{cases} ∠ ADC = ∠ F, \\ ∠ ACD = ∠ CBF = 90°, \\ AC = CB, \end{cases}$ $\therefore △ ACD ≌ △ CBF(\mathrm{AAS}).$
(2) $\because △ ACD ≌ △ CBF, \therefore CD = BF. \because D$ 为 $BC$ 的中点, $\therefore CD = BD, \therefore BF = CD = BD = \dfrac{1}{2}BC = \dfrac{1}{2}AC,$ 则 $AC = 2BF.$

5. 如图,在$△ ABC$中,$∠ C=90°$,点$D$在边$AC$上,过点$A$作$AE ⊥ AB$交$BD$的延长线于$E$,过点$E$作$EM ⊥ AC$于$M$,且$AE=AD$,$∠ AED=∠ ADE$.
(1)求证:$BE$平分$∠ CBA$;
(2)求证:$AB=EM+BC$.
＞＞ 对点专练 P39

＞＞ 根据诊断结果 请完成对应的练习

5. (1) 如图, $\because ∠ C = 90°, \therefore ∠ 3 + ∠ BDC = 90°. \because ∠ BDC = ∠ ADE, \therefore ∠ 3 + ∠ ADE = 90°. \because ∠ EAB = 90°, \therefore ∠ 4 + ∠ AED = 90°. \because ∠ AED = ∠ ADE, \therefore ∠ 3 = ∠ 4, \therefore BE$ 平分 $∠ CBA.$

(2) 如图,过点 $D$ 作 $DF ⊥ AB$ 于点 $F, \because EM ⊥ AC, \therefore ∠ AME = ∠ AFD = 90°, \therefore ∠ 2 + ∠ CAE = 90°. \because AE ⊥ AB, \therefore ∠ EAB = 90°, \therefore ∠ 1 + ∠ CAE = 90°, \therefore ∠ 1 = ∠ 2$. 在 $△ AME$ 与 $△ DFA$ 中,
$\begin{cases} ∠ AME = ∠ DFA, \\ ∠ 2 = ∠ 1, \\ AE = DA, \end{cases}$ $\therefore △ AME ≌ △ DFA (\mathrm{AAS}), \therefore EM = AF.$
$\because DF ⊥ AB, \therefore ∠ BFD = 90°. \because ∠ C = 90°, \therefore ∠ BFD = ∠ C.$ 由 (1) 可知, $∠ 3 = ∠ 4$, 在 $△ BFD$ 和 $△ BCD$ 中, $\begin{cases} ∠ BFD = ∠ C, \\ ∠ 4 = ∠ 3, \\ BD = BD, \end{cases}$
$\therefore △ BFD ≌ △ BCD(\mathrm{AAS}), \therefore BC = BF, \therefore AB = AF + BF = EM + BC.$