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8. (2025·盘龙区模拟)如图，∠1=∠2，∠A=∠B，AE=BE，点D在边AC上，AE与BD相交于点O. 求证:△AEC≌△BED. 
答案
8. 证明: $\because ∠1 = ∠2, \therefore ∠1 + ∠AED = ∠2 + ∠AED,$即$∠AEC=∠BED$,在$△ AEC$ 和$△ BED$ 中,
$\begin{cases}∠A=∠B,\\AE=BE,\\∠AEC=∠BED,\end{cases}$
$\therefore △ AEC ≌ △ BED (\mathrm{ASA}).$
$\begin{cases}∠A=∠B,\\AE=BE,\\∠AEC=∠BED,\end{cases}$
$\therefore △ AEC ≌ △ BED (\mathrm{ASA}).$
9. (2024秋·潮阳区期末)如图，$AC=AE$，$∠ C=∠ E$，$∠ 1=∠ 2$.求证:$△ ABC≌△ ADE$.
答案
9. 证明: $\because ∠1 = ∠2, \therefore ∠1 + ∠EAC = ∠2 + ∠EAC,\therefore ∠BAC = ∠DAE$,在$△ ABC$ 和$△ ADE$ 中
$\begin{cases}∠BAC=∠DAE,\\AC=AE,\\∠C=∠E,\end{cases}$
$\therefore △ ABC ≌ △ ADE (\mathrm{ASA}).$
$\begin{cases}∠BAC=∠DAE,\\AC=AE,\\∠C=∠E,\end{cases}$
$\therefore △ ABC ≌ △ ADE (\mathrm{ASA}).$
10. (2024秋·房山区期末)如图,点B是线段AD上一点,BC//DE,AB=ED,∠A=∠E.求证:△ABC≌△EDB.
答案
10. 证明:$\because BC// DE,\therefore ∠ABC=∠D$,在$△ ABC$ 和$△ EDB$ 中, $\begin{cases}∠A=∠E,\\AB=ED,\\∠ABC=∠D,\end{cases}$
$\therefore △ ABC ≌ △ EDB (\mathrm{ASA}).$
$\therefore △ ABC ≌ △ EDB (\mathrm{ASA}).$
11. (2024·鼓楼区模拟)如图,在四边形ABCD中,AB//CD,在BD上取两点E,F,使DF=BE,连接AE,CF.若AE//CF,试说明△ABE≌△CDF. 
答案
11. $\because AB// CD,\therefore ∠ABE = ∠CDF.$ $\because AE// CF,\therefore ∠AEB=∠CFD$,在$△ ABE$ 和$△ CDF$ 中,$\begin{cases}∠ABE=∠CDF,\\BE=DF,\\∠AEB=∠CFD,\end{cases}$
$\therefore △ ABE≌△ CDF(\mathrm{ASA}).$
$\therefore △ ABE≌△ CDF(\mathrm{ASA}).$
