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9. 如图8，一块大的三角板ABC，D是AB上一点，现要求过点D割出一块小的三角板ADE，使$DE// BC$，请作出DE.（不写作法，保留作图痕迹）
答案
9. 解：如答图1，DE即为所作.
10. 已知点O为直线AB上一点，将直角三角板MON的直角顶点放在点O上，并在∠MON内部作射线OC.
（1）如图9①，三角板的一边OM与射线OA重合，则∠AOC的余角是
（2）将三角板按照如图9②的方式放置，仅满足OC平分∠MOB，求∠AOM与∠NOC之间的数量关系.
（3）若仍将三角板按照如图9②所示的方式放置，使OC恰好平分∠MOB，且∠BON = 4∠NOC，求∠AOM的度数.
（1）如图9①，三角板的一边OM与射线OA重合，则∠AOC的余角是
$∠CON$
，补角是$∠COB$
.（2）将三角板按照如图9②的方式放置，仅满足OC平分∠MOB，求∠AOM与∠NOC之间的数量关系.
（3）若仍将三角板按照如图9②所示的方式放置，使OC恰好平分∠MOB，且∠BON = 4∠NOC，求∠AOM的度数.
答案
10. (1) $∠CON$ $∠COB$
(2) $\because OC$平分$∠MOB$，
$\therefore ∠COM = ∠COB = ∠NOC + ∠NOB$.
$\because ∠MON = 90°$，$\therefore ∠AOM + ∠NOB = 180° - ∠MON = 90°$.
$\therefore ∠AOM + ∠NOB = ∠MON = ∠COM + ∠NOC$.
$\therefore ∠AOM + ∠NOB = ∠NOC + ∠NOB + ∠NOC$.
$\therefore ∠AOM = 2∠NOC$.
(3) 设$∠NOC = x$，则$∠BON = 4∠NOC = 4x$.
$\because OC$恰好平分$∠MOB$，$\therefore ∠COM = ∠COB = ∠NOC + ∠BON = 5x$.
又$\because ∠MON = 90°$，即$∠COM + ∠NOC = 90°$，$\therefore 5x + x = 90°$. 解得$x = 15°$.
$\therefore ∠NOC = 15°$，由 (2) 可得，$∠AOM = 2∠NOC = 30°$.
(2) $\because OC$平分$∠MOB$，
$\therefore ∠COM = ∠COB = ∠NOC + ∠NOB$.
$\because ∠MON = 90°$，$\therefore ∠AOM + ∠NOB = 180° - ∠MON = 90°$.
$\therefore ∠AOM + ∠NOB = ∠MON = ∠COM + ∠NOC$.
$\therefore ∠AOM + ∠NOB = ∠NOC + ∠NOB + ∠NOC$.
$\therefore ∠AOM = 2∠NOC$.
(3) 设$∠NOC = x$，则$∠BON = 4∠NOC = 4x$.
$\because OC$恰好平分$∠MOB$，$\therefore ∠COM = ∠COB = ∠NOC + ∠BON = 5x$.
又$\because ∠MON = 90°$，即$∠COM + ∠NOC = 90°$，$\therefore 5x + x = 90°$. 解得$x = 15°$.
$\therefore ∠NOC = 15°$，由 (2) 可得，$∠AOM = 2∠NOC = 30°$.
11. 如图10，已知E，F为四边形ABDC的边CA的延长线上的两点，连接DE，BF，作∠BDH的平分线DP交AB的延长线于点P. 若∠1=∠2，∠3=∠4，∠5=∠C.
（1）判断DE与BF是否平行，并说明理由；
（2）试说明：∠C=2∠P.
（1）判断DE与BF是否平行，并说明理由；
（2）试说明：∠C=2∠P.
答案
11. 解：(1) $DE// BF$.
理由如下：$\because ∠3 = ∠4$，根据“内错角相等，两直线平行”，$\therefore BD// CE$.
根据“两直线平行，内错角相等”，$\therefore ∠5 = ∠FAB$.
$\because ∠5 = ∠C$，$\therefore ∠C = ∠FAB$. 根据“同位角相等，两直线平行”，$\therefore AB// CD$.
根据“两直线平行，内错角相等”，$\therefore ∠2 = ∠BGD$.
$\because ∠1 = ∠2$，$\therefore ∠1 = ∠BGD$. 根据“内错角相等，两直行平行”，$\therefore DE// BF$.
(2) $\because AB// CD$，根据“两直线平行，内错角相等”，$\therefore ∠P = ∠PDH$.
$\because DP$平分$∠BDH$，$\therefore ∠BDP = ∠PDH$. $\therefore ∠BDP = ∠PDH = ∠P$.
$\because ∠5 = ∠BDH = ∠PDH + ∠BDP$，$\therefore ∠5 = 2∠P$.
$\because ∠C = ∠5$，$\therefore ∠C = 2∠P$.
理由如下：$\because ∠3 = ∠4$，根据“内错角相等，两直线平行”，$\therefore BD// CE$.
根据“两直线平行，内错角相等”，$\therefore ∠5 = ∠FAB$.
$\because ∠5 = ∠C$，$\therefore ∠C = ∠FAB$. 根据“同位角相等，两直线平行”，$\therefore AB// CD$.
根据“两直线平行，内错角相等”，$\therefore ∠2 = ∠BGD$.
$\because ∠1 = ∠2$，$\therefore ∠1 = ∠BGD$. 根据“内错角相等，两直行平行”，$\therefore DE// BF$.
(2) $\because AB// CD$，根据“两直线平行，内错角相等”，$\therefore ∠P = ∠PDH$.
$\because DP$平分$∠BDH$，$\therefore ∠BDP = ∠PDH$. $\therefore ∠BDP = ∠PDH = ∠P$.
$\because ∠5 = ∠BDH = ∠PDH + ∠BDP$，$\therefore ∠5 = 2∠P$.
$\because ∠C = ∠5$，$\therefore ∠C = 2∠P$.
