8. 如图18-10,以$BC为底边的等腰三角形ABC$,点$D$,$E$,$G分别在BC$,$AB$,$AC$上,且$EG// BC$,$DE// AC$,延长$GE至点F$,使得$BE = BF$.

(1)求证：四边形$BDEF$为平行四边形；
(2)当$∠C = 45^{\circ}$,$BD = 2$时,求$D$,$F$两点间的距离为.

(1)∵ △ABC是以BC为底边的等腰三角形，∴ ∠ABC = ∠C。∵ EG//BC，DE//AC，∴ ∠AEG = ∠ABC = ∠C，四边形CDEG是平行四边形，∴ ∠DEG = ∠C。∵ BE = BF，∴ ∠BFE = ∠BEF。又∵ ∠BEF = ∠AEG = ∠ABC，∴ ∠BFE = ∠DEG，∴ BF//DE，∴ 四边形BDEF为平行四边形。 (2)∵ ∠C = 45°，∴ ∠ABC = ∠BFE = ∠BEF = 45°，∴ △BDE，△BEF是等腰直角三角形，∴ BF = BE = $\frac{\sqrt{2}}{2}$BD = $\sqrt{2}$。作FM⊥BD于点M，连接DF，则△BFM是等腰直角三角形，∴ FM = BM = $\frac{\sqrt{2}}{2}$BF = 1，∴ DM = 3。在Rt△DFM中，由勾股定理，得DF = $\sqrt{1^{2} + 3^{2}}$ = $\sqrt{10}$，即D，F两点间的距离为$\sqrt{10}$。

9. 如图18-11,在$\triangle ABC$中,$AB = AC$,点$D在BC$上,以$AD$,$AE为腰作等腰三角形ADE$,且$∠ADE = ∠ABC$. 连接$CE$,过点$E作EM// BC交CA的延长线于点M$,连接$BM$.
(1)求证：$\triangle BAD\cong \triangle CAE$；

(2)若$∠ABC = 30^{\circ}$,求$∠MEC$的度数；

(3)求证：四边形$MBDE$是平行四边形.

(1)∵ AB = AC，∴ ∠ABC = ∠ACB，∴ ∠BAC = 180° - 2∠ABC。∵ 以AD，AE为腰作等腰三角形ADE，∴ AD = AE，∴ ∠ADE = ∠AED，∴ ∠DAE = 180° - 2∠ADE。∵ ∠ADE = ∠ABC，∴ ∠BAC = ∠DAE，∴ ∠BAC - ∠CAD = ∠DAE - ∠CAD，∴ ∠BAD = ∠CAE。在△BAD和△CAE中，$\left\{\begin{array}{l}AB = AC,\\\angle BAD = \angle CAE,\\AD = AE,\end{array}\right.$ ∴ △BAD ≌ △CAE(SAS)。 (2)∵ AB = AC，∴ ∠ACB = ∠ABC = 30°。∵ △BAD ≌ △CAE，∴ ∠ABD = ∠ACE = 30°，∴ ∠ACB = ∠ACE = 30°，∴ ∠ECB = ∠ACB + ∠ACE = 60°。∵ EM//BC，∴ ∠MEC + ∠ECD = 180°，∴ ∠MEC = 180° - 60° = 120°。 (3)∵ △BAD ≌ △CAE，∴ DB = EC，∠ABD = ∠ACE。∵ AB = AC，∴ ∠ABD = ∠ACB，∴ ∠ACB = ∠ACE。∵ EM//BC，∴ ∠EMC = ∠ACB，∴ ∠ACE = ∠EMC，∴ ME = EC，∴ DB = ME。又∵ EM//BD，∴ 四边形MBDE是平行四边形。