8. 如图,在$△ ABC$中,点D在边BC上,$CD=AB$, $DE// AB$, $∠ DCE=∠ A$. 求证:$DE=BC$. (第8题图)

8. $\because \quad DE// AB$,
$\therefore \quad ∠ EDC= ∠ B$.
在$△ CDE$ 和$△ ABC$ 中,
$\begin{cases}∠ EDC = ∠ B, \\CD = AB, \\∠ DCE = ∠ A,\end{cases}$
$\therefore \quad △ CDE ≌ △ ABC(\mathrm{ASA})$.
$\therefore \quad DE=BC$.

9. 如图,在$△ ABC$中,点$D$在$AB$上,点$E$在$BC$上,且$BD=BE$.
(1)请你再添加一个条件,使得$△ BEA ≌ △ BDC$,并说明理由,你添加的条件是;依据是.
(2)根据你添加的条件,再写出图中的一对全等三角形,并说明理由.

(第9题图)

9. (1) $∠ BEA = ∠ BDC$ ASA
(2) $△ DFA ≌ △ EFC$.
理由:$\because \quad △ BEA ≌ △ BDC$,
$\therefore \quad ∠ DAF = ∠ ECF, AB = CB$.
$\because \quad BD = BE$,
$\therefore \quad AB - BD = CB - BE,即 AD = CE$.
在$△ DFA$ 和$△ EFC$ 中,
$\begin{cases}∠ DFA = ∠ EFC, \\∠ DAF = ∠ ECF, \\AD = CE,\end{cases}$
$\therefore \quad △ DFA ≌ △ EFC(\mathrm{AAS})$.

10. 如图，M是线段AB上一点，ED是过点M的一条线段，连结AE，BD，过点B作BF//AE，交ED于点F，且EM=FM.
(1)若AE=5，求BF的长.
(2)若∠AEC=90°，∠DBF=∠CAE，求证：CD=FE.

(第10题图)

10. (1)$\because \quad BF// AE$,
$\therefore \quad ∠ EAM = ∠ FBM, ∠ E = ∠ BFM$.
在$△ AEM$ 和$△ BFM$ 中,
$\begin{cases}∠ EAM = ∠ FBM, \\∠ E = ∠ BFM, \\EM = FM,\end{cases}$
$\therefore \quad △ AEM ≌ △ BFM(\mathrm{AAS})$.
$\therefore \quad AE = BF$.
$\because \quad AE = 5$,
$\therefore \quad BF = 5$.
(2)$\because \quad BF// AE$,
$\therefore \quad ∠ AEC = ∠ BFM$.
$\because \quad ∠ AEC = 90°$,
$\therefore \quad ∠ BFM = 90°$.
$\therefore \quad ∠ BFD = 180° - 90° = 90°$.
$\therefore \quad ∠ AEC = ∠ BFD$.
由(1)知 $AE = BF$,
在$△ ACE$ 和$△ BDF$ 中,
$\begin{cases}∠ CAE = ∠ DBF, \\AE = BF, \\∠ AEC = ∠ BFD,\end{cases}$
$\therefore \quad △ ACE ≌ △ BDF(\mathrm{ASA})$.
$\therefore \quad CE = DF$.
$\therefore \quad DF - CF = CE - CF$,
即 $CD = FE$.