4. 如图,点 D 在线段 BC 上,∠B = ∠C = ∠ADE=60°,AB=DC.
(1)求证:△ABD≌△DCE.
(2)判断△ADE 是什么特殊三角形,并说明理由.

(第4题图)

4. (1)$\because ∠ ADC = ∠ ADE+∠ CDE=60°+∠ CDE,∠ ADC = ∠ B+∠ BAD=60°+∠ BAD$,
$\therefore ∠ BAD=∠ CDE$.
在$△ ABD$和$△ DCE$中,
$\begin{cases}∠ B=∠ C,\\AB=DC,\\∠ BAD=∠ CDE,\end{cases}$
$\therefore △ ABD ≌ △ DCE(\mathrm{ASA})$.
(2)$△ ADE$是等边三角形.
理由:由(1)知$△ ABD ≌ △ DCE$,
$\therefore AD=DE$.
又$\because ∠ ADE=60°$,
$\therefore △ ADE$是等边三角形.

5. 如图,在$△ ABC$中,$AB=AC$,点D在$△ ABC$内部,连结DB、DC、DA,$BD=BC$,$∠ DBC=60°$,点E在$△ ABC$外部,连结EA、EB、EC,$∠ BCE=150°$,$∠ ABE=60°$.
(1)求$∠ ADC$的度数.
(2)判断$△ ABE$的形状并加以证明.

(第5题图)

5. (1)$\because BD=BC,∠ DBC=60°$,
$\therefore △ DBC$是等边三角形.
$\therefore ∠ BDC=60°$.
在$△ ADB$和$△ ADC$中,
$\begin{cases}AB=AC,\\AD=AD,\\DB=DC,\end{cases}$
$\therefore △ ADB ≌ △ ADC(\mathrm{SSS})$.
$\therefore ∠ ADB=∠ ADC$.
$\therefore ∠ ADC = ∠ ADB=\frac{1}{2}(360°-∠ BDC)$
$=150°$.
(2)$△ ABE$是等边三角形.
证明:$\because ∠ ABE=∠ DBC=60°$,
$\therefore ∠ ABD=∠ CBE$.
在$△ ABD$和$△ EBC$中,
$\begin{cases}∠ ADB=∠ ECB=150°,\\BD=BC,\\∠ ABD=∠ EBC,\end{cases}$
$\therefore △ ABD ≌ △ EBC(\mathrm{ASA})$.
$\therefore AB=BE$.
$\because ∠ ABE=60°$,
$\therefore △ ABE$是等边三角形.

6. 已知在$△ ABC$中，$AB=AC$，$D$是边$AB$上一点，$∠ BCD=∠ A$.
(1)如图1，试说明$CD=CB$.
(2)如图2，过点$B$作$BE ⊥ AC$，垂足为点$E$，$BE$与$CD$相交于点$F$.
①试说明$∠ BCD=2∠ CBE$；
②如果$△ BDF$是等腰三角形，求$∠ A$的度数.

(第6题图)

6. (1)$\because AB=AC$,
$\therefore ∠ ABC=∠ ACB$.
$\because ∠ BDC$是$△ ADC$的一个外角,
$\therefore ∠ BDC=∠ A+∠ ACD$.
$\because ∠ ACB=∠ BCD+∠ ACD,∠ BCD=∠ A$,
$\therefore ∠ BDC=∠ ACB$.
$\therefore ∠ ABC=∠ BDC$.
$\therefore CD=CB$.
(2)①$\because BE⊥ AC$,
$\therefore ∠ BEC=90°$.
$\therefore ∠ CBE+∠ ACB=90°$.
设$∠ CBE=α$,则$∠ ACB=90°-α$.
$\therefore ∠ ACB=∠ ABC=∠ BDC=90°-α$.
$\therefore ∠ BCD = 180°-∠ BDC-∠ ABC$
$=180°-(90°-α)-(90°-α)$
$=2α$.
$\therefore ∠ BCD=2∠ CBE$.
②$\because ∠ BFD$是$△ CBF$的一个外角,
$\therefore ∠ BFD=∠ CBE+∠ BCD=α+2α=3α$.
分三种情况:
a. 当$BD=BF$时,
$\therefore ∠ BDC=∠ BFD=3α$.
$\because ∠ ACB=∠ ABC=∠ BDC=90°-α$,
$\therefore 90°-α=3α$.
$\therefore α=22.5°$.
$\therefore ∠ A=∠ BCD=2α=45°$.
b. 当$DB=DF$时,
$\therefore ∠ DBE=∠ BFD=3α$.
$\because ∠ DBE = ∠ ABC-∠ CBE=90°-α-α$
$=90°-2α$,
$\therefore 90°-2α=3α$.
$\therefore α=18°$.
$\therefore ∠ A=∠ BCD=2α=36°$.
c. 当$FB=FD$时,
$\therefore ∠ DBE=∠ BDF$.
$\because ∠ BDF=∠ ABC＞∠ DBF$,
$\therefore$ 不存在$FB=FD$.
综上所述,如果$△ BDF$是等腰三角形,那么$∠ A$的度数为$45°$或$36°$.