1. 如图, 在 $△ ABC$ 中, $∠ BAC = 60°$, $∠ C = 40°$, $P, Q$ 分别在 $BC, CA$ 上, 并且 $AP, BQ$ 分别是 $∠ BAC, ∠ ABC$ 的平分线.求证: $BQ+AQ=AB+BP$.

过点 P 作 $PD// BQ$ 交 $CQ$ 于点 D. $\because ∠ BAC=60°$, $∠ C=40°,\therefore ∠ ABC=180°-40°-60°=80°$.
$\because ∠ CBQ=\frac{1}{2}∠ ABC = \frac{1}{2} × 80° = 40°$,
$\therefore ∠ CBQ=∠ ACB,\therefore BQ=CQ,\therefore BQ+AQ = CQ+AQ = AC$ ①. $\because PD // BQ,\therefore ∠ CPD = ∠ CBQ=40°,\therefore ∠ CPD=∠ ACB=40°,\therefore PD= CD,∠ ADP=∠ CPD+∠ ACB = 40°+40° = 80°.\because ∠ ABC=80°$,
$\therefore ∠ ABC=∠ ADP.\because AP$ 平分 $∠ BAC$, $\therefore ∠ BAP = ∠ CAP$. 在$△ ABP$ 与 $△ ADP$ 中, $\begin{cases} ∠ ABP=∠ ADP, \\ ∠ BAP=∠ DAP, \\ AP=AP, \end{cases}$ $\therefore △ ABP ≌ △ ADP$(AAS),$\therefore AB = AD, BP = PD,\therefore AB+BP = AD+PD = AD+CD = AC$ ②.由①②可得,$BQ+AQ=AB+BP$.

2. 如图,在$△ ABC$中,AD平分$∠ BAC$交BC边于点D,点E是BC边的中点,线段$EF // AD$交线段AB于点G,交线段CA的延长线于点F.
(1)若$CF=6,AG=2$,求AC的长;
(2)求证:$BG=CF$.

(1)$\because AD$ 平分 $∠ BAC$, $\therefore ∠ BAD = ∠ DAC. \because AD // EF$,
$\therefore ∠ DAC = ∠ F, ∠ BAD = ∠ FGA, \therefore ∠ F = ∠ FGA, \therefore AG = AF$.
$\because CF=6,AG=2,\therefore AC=CF-AF=CF-AG=6-2=4$.
(2)如图,作 $CM// AB$ 交 $FE$ 的延长线于点 $M.\because BG // CM,\therefore ∠ B = ∠ MCE.\because E$ 是 BC的中点,$\therefore BE = EC$. 在 $△ BEG$ 和 $△ CEM$ 中,$\begin{cases} ∠ B=∠ MCE, \\ BE=CE, \\ ∠ BEG=∠ CEM, \end{cases}$
$\therefore △ BEG ≌ △ CEM$(ASA),$\therefore BG = CM. \because AD // EF, \therefore ∠ 1 = ∠ FGA,∠ 2 = ∠ F. \because ∠ 1 = ∠ 2, \therefore ∠ F = ∠ FGA. \because AB // CM$,
$\therefore ∠ FGA = ∠ M,\therefore ∠ F = ∠ M,\therefore CF = CM,\therefore BG = CF$.

3. （2026·杭州期中）如图，AD为$△ ABC$的角平分线，$CE ⊥ AD$交$AD$的延长线于点$E$，$∠ BAD=2∠ DCE=2α$。
（1）求证：$△ ABD$为等腰三角形；
（2）若$DA=DC$，$BD=4$，求$DE$的长；
（3）在（2）的条件下，求证：$AD+AC=2AE$。

(1)$\because ∠ BAD=2∠ DCE=2α,\therefore ∠ DCE=α.\because CE⊥ AD$ 交 AD的延长线于点 E,$\therefore ∠ E=90°,\therefore ∠ CDE=90°-α,\therefore ∠ ADB=∠ CDE=90°-α$. 在 $△ ABD$ 中, $∠ ABD+∠ ADB+∠ BAD = 180°$,
$\therefore ∠ ABD=180°-∠ BAD-∠ ADB=180°-2α-(90°-α)=90°-α$,
$\therefore ∠ ABD=∠ ADB,\therefore AB=AD,\therefore △ ABD$ 为等腰三角形.
(2)如图①所示,过点 A 作 $AF⊥ BD$,由(1)可知$∠ ABD=90°-α$,$\because AD$ 为 $△ ABC$ 的角平分线, $\therefore ∠ BAD = ∠ DAC = 2α$,
$\therefore ∠ BAC=4α.\because DA=DC,\therefore ∠ DAC=∠ DCA=2α$. 在 $△ ABC$ 中,
$∠ ABC+∠ BAC+∠ ACB = 180°$,$\therefore 90°-α+4α+2α = 180°$,解得$α=18°$,$\therefore ∠ ABC=90°-α=72°$,$∠ BAC=4α=72°$,$∠ ACB=2α=36°$,$\therefore ∠ BAD=∠ DAC=36°$,$\therefore ∠ ADB=∠ DAC+∠ ACB=72°$,$\therefore ∠ ABC=∠ ADB$,$\therefore AB=AD.\because BD=4$,$\therefore DF = \frac{1}{2}BD = \frac{1}{2}×4=2$.
在 $△ ADF$ 和 $△ CDE$ 中, $\begin{cases} ∠ AFD=∠ CED, \\ ∠ ADF=∠ CDE, \\ AD=CD, \end{cases}$
$\therefore △ ADF ≌ △ CDE$(AAS),$\therefore DE=DF=2$.

(3)如图②所示,过点 C 作 $CM// AB$,$\therefore ∠ BAD=∠ M.\because ∠ BAD=∠ CAD=∠ ACD$,$\therefore ∠ ACB=∠ M$. 由(2)可知 $AB=AD=DC$,$∠ ABC=∠ ADB=∠ CDE=∠ BAC$, 在 $△ MDC$ 和 $△ CAB$ 中,$\begin{cases} ∠ M=∠ ACB, \\ ∠ CDM=∠ BAC, \\ CD=BA, \end{cases}$
$\therefore △ MDC≌△ CAB$(AAS),$\therefore AC=DM$,$\therefore AM=AD+DM=AD+AC=AE+ME$,$\therefore ∠ M=∠ ACB=∠ CAD$,$\therefore CA=CM.\because CE⊥ AM$,$\therefore AM=2AE$,$\therefore AD+AC=2AE$.