19. (6分)計(jì)算：(1)$180^{\circ}-(75^{\circ}44^{\prime}8^{\prime\prime}+14^{\circ}15^{\prime}52^{\prime\prime})$；(2)$12^{\circ}15^{\prime}38^{\prime\prime}×2 + 62^{\circ}17^{\prime}÷3$.
答案：解:(1)原式=180°?89°59′60″=180°?90°=90°. (2)原式=24°30′76″+20°+(137÷3)′=44°30′76″+45′+120″÷3 =44°75′76″+40″=45°16′56″.
解析：
(1)解:原式=180°?(75°44′8″+14°15′52″)
=180°?89°59′60″
=180°?90°
=90°.
(2)解:原式=(12°×2+15′×2+38″×2)+(62°÷3+17′÷3)
=24°30′76″+20°+(137′÷3)
=24°30′76″+20°45′40″
=44°75′116″
=45°16′56″.
20. (6分)尺規(guī)作圖：如圖,在平面內(nèi)有A,B,C三點(diǎn).
(1)畫(huà)直線(xiàn)AB,射線(xiàn)AC,線(xiàn)段BC.
(2)在線(xiàn)段BC上任取一點(diǎn)D(不同于點(diǎn)B,C),連接AD,并延長(zhǎng)AD至點(diǎn)E,使$DE = AD$.
答案：解:(1)如答圖，直線(xiàn)AB,射線(xiàn)AC,線(xiàn)段BC即為所求. (2)如答圖，線(xiàn)段AD,DE即為所求.
21. (8分)如圖,$\angle AOB$是平角,OM,ON分別是$\angle AOC$,$\angle BOD$的平分線(xiàn).
(1)若$\angle AOC = 40^{\circ}$,$\angle BOD = 60^{\circ}$,求$\angle MON$的度數(shù).
(2)若只給條件“$\angle COD = 80^{\circ}$”,你能求出$\angle MON$的度數(shù)嗎？如果能,請(qǐng)給出過(guò)程；如果不能,請(qǐng)說(shuō)明理由.
答案：解:(1)∵∠AOB是平角,∠AOC=40°,∠BOD=60°,∴∠COD=∠AOB?∠AOC?∠BOD=180°?40°?60°=80°.∵OM,ON分別是∠AOC,∠BOD的平分線(xiàn)，∴∠MOC=$\frac{1}{2}$∠AOC=20°,∠NOD=$\frac{1}{2}$∠BOD=30°.∴∠MON=∠MOC+∠COD+∠NOD=20°+80°+30°=130°. (2)能.過(guò)程如下:∵OM,ON分別是∠AOC,∠BOD的平分線(xiàn)，∴∠MOC+∠NOD=$\frac{1}{2}$∠AOC+$\frac{1}{2}$∠BOD=$\frac{1}{2}$(∠AOC+∠BOD)=$\frac{1}{2}$×(180°?80°)=50°.∴∠MON=∠MOC+∠NOD+∠COD=50°+80°=130°.
22. (8分)如圖,已知$AB= \frac{1}{4}BM= \frac{1}{5}AN$,點(diǎn)C,D分別是BM,AN的中點(diǎn),且$CD = 14$.
(1)求證：$BM = BN$；
(2)求MN的長(zhǎng).
答案：(1)證明:因?yàn)锳B=$\frac{1}{4}$BM=$\frac{1}{5}$AN,所以BM=4AB,AN=5AB,則BN=AN?AB=4AB,所以BM=BN. (2)解:設(shè)AB=x,則BM=4x,AN=5x,所以BN=4x,MN=BM+BN=8x.又因?yàn)镃,D分別是BM,AN的中點(diǎn)，所以CD=BC+BD=BC+(AD?AB)=$\frac{1}{2}$BM+($\frac{1}{2}$AN?AB),所以2x+(2.5x?x)=14,解得x=4,則MN=8x=32.
解析：
(1)證明：因?yàn)?AB = \frac{1}{4}BM = \frac{1}{5}AN$，所以$BM = 4AB$，$AN = 5AB$，則$BN = AN - AB = 5AB - AB = 4AB$，所以$BM = BN$。
(2)解：設(shè)$AB = x$，則$BM = 4x$，$AN = 5x$，由(1)知$BN = 4x$，所以$MN = BM + BN = 4x + 4x = 8x$。因?yàn)辄c(diǎn)$C$，$D$分別是$BM$，$AN$的中點(diǎn)，所以$BC = \frac{1}{2}BM = 2x$，$AD = \frac{1}{2}AN = 2.5x$，則$BD = AD - AB = 2.5x - x = 1.5x$，所以$CD = BC + BD = 2x + 1.5x = 3.5x$。又因?yàn)?CD = 14$，所以$3.5x = 14$，解得$x = 4$，則$MN = 8x = 8×4 = 32$。
