1. 如图,在 $\mathrm{Rt}△ ABC$ 中, $∠ BAC=90°,AB=AC,D$ 是 $△ ABC$ 外一点,且 $AD=AC$.求 $∠ BDC$ 的度数.

1. 设 $∠ ADB = α, \because AB = AC = AD, \therefore ∠ ABD = ∠ ADB = α$,$\therefore ∠ BAD=180°-(∠ ABD+∠ ADB)=180°-2α, \therefore ∠ CAD=∠ BAD-∠ BAC=180°-2α-90°=90°-2α.\because AD=AC,\therefore ∠ ADC=∠ ACD=\frac{180°-∠ CAD}{2}=\frac{180°-(90°-2α)}{2}=45°+α,\therefore ∠ BDC=∠ ADC-∠ ADB=45°+α-α=45°.$

2. 如图,已知$AB=AC$,$∠ BAC=2∠ BDC$,求证:$AD=AB$.

2. 如图,设 $∠ BDC = α$, 则 $∠ BAC = 2α$, 作$∠ EAD = ∠ BAC = 2α$, 截取 $AE = AD$, 连接$CE, DE$, 由题可求得 $△ ABD ≌ △ ACE$(SAS), $\therefore BD = CE, ∠ AEC = ∠ ADB$, 在$△ AEF$ 和 $△ CFD$ 中, $∠ DAE + ∠ AEF + ∠ AFE = 180°, ∠ ADC + ∠ ECD + ∠ CFD = 180°, \therefore ∠ DAE+∠ AEF = ∠ ADC+∠ ECD$,$\therefore ∠ ECD=α,\therefore △ BCD≌△ EDC(\mathrm{SAS}),\therefore BC=DE,\therefore △ BAC≌△ DAE(\mathrm{AAS}),\therefore AD=AB.$

3. 如图,在等边$△ ABC$中,将线段$AC$绕点$A$顺时针旋转$α(0°＜α＜60°)$,得到线段$AD$,连接$CD$,作$∠ BAD$的平分线$AE$,交$BC$于点$E$.
(1)①根据题意,补全图形;
②请用等式写出$∠ BAD$与$∠ BCD$的数量关系,并证明.
(2)分别延长$CD$和$AE$交于点$F$,用等式表示线段$AF$,$CF$,$DF$的数量关系,并证明.

3. (1)①补全图形,如图①所示.
②$2 ∠ BCD = ∠ BAD$. 证明如下: $\because △ ABC$ 是等边三角形,$\therefore ∠ BAC= ∠ ACB=60°.\because$ 将线段 $AC$ 绕点 $A$ 顺时针旋转$α(0°＜α＜60°)$, 得到线段 $AD, \therefore AC = AD, ∠ CAD = α, \therefore ∠ ACD=∠ ADC=\frac{180°-α}{2},\therefore ∠ BAD = 60°-α, ∠ BCD = \frac{180°-α}{2}-60° = 30°-\frac{α}{2},\therefore 2∠ BCD=∠ BAD.$
(2)$AF=CF+DF$. 证明如下: 如图②, 过点 $A$ 作 $AH ⊥ CD$ 于点 $H$.
$\because AF$ 平分 $∠ BAD, \therefore ∠ DAF = 30°-\frac{α}{2}, \therefore ∠ F = ∠ ADC - ∠ DAF=\frac{180°-α}{2}-(30°-\frac{α}{2})=60°.\because AH ⊥ CD, \therefore ∠ DAH=∠ CAH,∠ FAH=30°,\therefore AF=2FH.\because AD=AC,AH ⊥ CD,\therefore DH=CH=\frac{1}{2}CD,\therefore AF=2FH=2(CF-CH)=2(CF-\frac{1}{2}CD)=2[CF-\frac{1}{2}(CF-DF)]=CF+DF.$