3. 如图,在$△ ABC$中,$CD$是$∠ ACB$的平分线,$BE ⊥ CD$,垂足为点$E$.
(1)若$∠ CBE - ∠ DBE = 40°$,求$∠ BAC$的度数;
(2)连接$AE$,若$△ ABC$的面积为$6$,求$△ AEC$的面积.

3. (1)延长 BE 交 AC 于点 F.
∵ CD 平分∠ACB,
∴ ∠FCE = ∠BCE.
∵ BE ⊥ CD,
∴ ∠BEC = ∠FEC = 90°.
∵ CE = CE,
∴ △BCE≌△FCE(ASA),
∴ ∠CBE = ∠CFE.
∵ ∠CFE = ∠DBE + ∠BAC,
∴ ∠CBE = ∠DBE + ∠BAC.
∵ ∠CBE - ∠DBE = 40°,
∴ ∠BAC=40°.
(2)由(1)可得,△BCE≌△FCE,
∴ BE = EF,
∴ $S_{△BEC} = S_{△FEC}$, $S_{△ABE} = S_{△AEF}$.
∵ $S_{△BEC} + S_{△FCE} + S_{△ABE} + S_{△AEF} = S_{△ABC} = 6$,
∴ $S_{△AEC} = S_{△AEF} + S_{△FCE} = \frac{1}{2}S_{△ABC} = 3$.

4. 如图,在$△ ABC$中,$AB=AC$,$∠ A=90°$,$BE$是$△ ABC$的角平分线,$CD ⊥ BE$交$BE$的延长线于点$D$. 求证:$BE=2CD$.

4. 如图,延长 BA 和 CD 交于 Q.
∵ ∠CAQ = ∠BAE = ∠BDC = 90°,
∴ ∠ACQ + ∠Q = 90°, ∠ABE + ∠Q = 90°,
∴ ∠ACQ = ∠ABE. 在△ABE 和△ACQ 中, $\begin{cases} ∠ABE=∠ACQ,\\ AB=AC,\\ ∠BAE=∠CAQ, \end{cases}$
∴ △ABE≌△ACQ(ASA),
∴ BE = CQ.
∵ BD 平分∠ABC,
∴ ∠QBD = ∠CBD.
∵ ∠BDC = 90°,
∴ ∠BDC = ∠BDQ = 90°. 在△QDB 和△CDB 中,
$\begin{cases} ∠QBD=∠CBD,\\ BD=BD,\\ ∠BDQ=∠BDC, \end{cases}$
∴ △QDB ≌ △CDB (ASA),
∴ DQ = CD,
∴ BE = CQ = 2CD.

5. 如图,在$△ ABC$中,$∠ ABC=∠ ACB=40°$,$BD$是$∠ ABC$的平分线,延长$BD$至$E$,使$DE=AD$,求$∠ ECA$的度数.

5. 如图,在 BC 上截取 BF=AB,连接 DF.
∵ BD 是∠ABC 的平分线,
∴ ∠ABD = ∠FBD = 20°. 在△ABD 和△FBD 中,
$\begin{cases} AB=FB,\\ ∠ABD=∠FBD,\\ BD=BD, \end{cases}$
∴ △ABD≌△FBD(SAS),
∴ DF = DA = DE. 又
∵ ∠ACB = ∠ABC = 40°,
∴ ∠DFB = ∠A = 100°,
∴ ∠DFC = 180° - ∠DFB = 80°,
∴ ∠FDC = 60°.
∵ ∠EDC = ∠ADB = 180° - ∠ABD - ∠A = 180° - 20° - 100° = 60°,
∴ ∠FDC = ∠EDC. 又
∵ DF = DE, DC = DC,
∴ △DCF≌△DCE (SAS),
∴ ∠ECA = ∠DCB = 40°.

6. 如图,在$△ ABC$中,$∠ B=60°$,$△ ABC$的角平分线$AD$,$CE$相交于点$O$.
(1)求$∠ AOC$的度数;
(2)求证:$AE+CD=AC$.

6. (1)在△ABC 中,∠B = 60°,
∴ ∠BAC + ∠BCA = 180° - ∠B = 180°-60° = 120°.
∵ AD 平分∠BAC,CE 平分∠ACB,
∴ ∠OAC = ∠OAB = $\frac{1}{2}$∠BAC, ∠OCD = ∠OCA = $\frac{1}{2}$∠ACB. 在△OAC 中,
∠AOC = 180° - (∠OAC + ∠OCA) = 180° - $\frac{1}{2}$(∠BAC + ∠BCA) = 180° - $\frac{1}{2}$×120° = 120°.
(2)
∵ ∠AOC = 120°,
∴ ∠AOE = ∠DOC = 180° - ∠AOC = 180° - 120° = 60°. 如图,在 AC 上截取 AF = AE,连接 OF, 在△AOE 和△AOF 中, $\begin{cases} AE=AF,\\ ∠OAE=∠OAF,\\ OA=OA, \end{cases}$
∴ △AOE≌△AOF(SAS),
∴ ∠AOE = ∠AOF,
∴ ∠AOF = 60°,
∴ ∠COF = ∠AOC - ∠AOF = 120° - 60° = 60°. 又
∵ ∠COD = ∠AOE = 60°,
∴ ∠COD = ∠COF. 在△COD 和△COF 中,
$\begin{cases} ∠COD=∠COF,\\ OC=OC,\\ ∠OCD=∠OCF, \end{cases}$
∴ △COD ≌ △COF (ASA),
∴ CD = CF. 又
∵ AF=AE,
∴ AC=AF+CF=AE+CD,即AE+CD=AC.