parachuted+back+to
答：在这儿，呵呵 Prison Break Episode 101 "Pilot"Airdate: 08/29/2005 Michael Scofield watches a tattoo artist add the finishing touches to a section of his arm. She marvels at her masterpiece. In just a matter of months, Michael has tattooed his chest, back and both arms down to...
答：5. The Metro journey back to the centre of the town was hot and uncomfortable.乘地铁返回市中心又热又不舒服。6. When travelling at bus or metro, please hold the handrail.坐公交或地铁时, 请抓紧扶手.7. B And there will be new metro lines and transport systems.而且到时将有新...
答：-Berwick: Name and back number -Michael: Scofield Michael 94941 Scofield Michael 94941。 -Berwick: Are you a religious man, Scofield? -Michael: ...Parachuted out of a plane 30 years ago with a million and a half in cash -Michael: Doesn't look like the type. -Fernando: Who does? Hey...
答：务期 reached the goal which annihilated the enemy. If oneflustered, one is lax, or slightly saves kernel of the woman, ownnamely its evil. The air fight and the ground warfare are same, thefight needs to be well-trained, enters may attack, draws back candefend, but also mus...
答：- Name and back number. Michael: Scofield, Michael. 94941. - Are you a religious(信奉宗教的) man, Scofield? Michael: Never really thought about...Cooper. Parachuted out of a plane 30 years ago with a million and a half in cash. Michael: Doesn't look like the type. Sucre: Who does?
答：下面是第一集的一部分，如果要更多，你可以看看http://www.maynet.cn/12385.html Prison Break Season 1 Episode 1 - That's it.- Can I just, you know, look at it for a minute?- You're an artist, Sid.- You're telling me you're just gonna walk out of here and I'm ...
答：Gerry and Fassbach fly to Camp Humphreys, a military base in South Korea, where they are attacked by zombies. Running back into the aircraft, Fassbach slips, falls and accidentally discharges his gun, killing himself. After being rescued by the base's surviving personnel, led by ...
答：5. The Metro journey back to the centre of the town was hot and unfortable.乘地铁返回市中心又热又不舒服。6. When travelling at bus or metro, please hold the handrail.坐公交或地铁时, 请抓紧扶手.7. B And there will be new metro lines and transport systems.而且到时将有新的...
答：The bodies of Wattana *** , 36, and Quach , 52, were found in a storage room in the back of the market.Quach had served with the French against the Vietnamese who were trying to expel the French coloniapsts and finally succeeded in 1954, Tourison said last week."' ...
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蔺娟15861378940：
三角形ABC中,P在射线BD上,且角BPC=角BAC(1)求证:PA平分角DPC,(2)若角BAC=60度,不可用圆和相似,还没学 -
10700余歪
： 作AF垂直PC,AE垂直BDAB=AC,角BPC=角BAC角BPC+角ABP+角ABC+角BCP=角BAC+角ABC+角ACP+角PCB化简得角ABP=角ACP因为AB=AC,角ABP=角ACP,两个直角,所以三角形AEB全等于三角形AFC,AE=AF所以PA平分角DPC答案补充 在PC上做AG=AP得因为角BAC=60度,易证角APC=60度所以PG=AG=APAC=AB,AG=AP,角GAC=角PAC-角PAG=角PAC-60度=角PAC-角BAC=角PAB所以三角形AGC全等于三角形APB所以GC=PBPC=PG+GC=AP+PB即PA+PB=PC
蔺娟15861378940：
已知,如图,AD是△ABD和△ACD的公共边.求证:∠BDC=∠BAC+∠B+∠C(用两种方法) -
10700余歪
：[答案] 证法1:∵在△ABD中,∠B+∠BAD+∠ADB=180°, 在△ACD中,∠C+∠ADC+∠CAD=180°, ∴∠ADB+∠ADC=360°-∠B-∠BAD-∠CAD-∠C=360°-∠B-∠BAC-∠C, ∵∠BDC=360°-(∠ADB+∠ADC)=∠BAC+∠B+∠C; 证法2:延长AD到E, ∵...
蔺娟15861378940：
如图,在三角形ABC中,AD是LBAC的平分线,L2=35,L4=65,求LADB的度数
10700余歪
： 按三角形内角和定理可知 角BAC+角2+角4=180 已知角2=35.角4=65 则角BAC=180-35-65=80 因为AD平分角BAC 所以角1=80/2=40 所以角ADB=180-40-35=105 度答案已给出! 满意请点下右下方的“采纳答案” 您的采纳是我的回答的动力!
蔺娟15861378940：
2)图⑵中,点P是△ABC外角平分线的交点,试探究∠BPC与∠A的关系. -
10700余歪
： 在AB的延长线、AC的延长线上分别任取点D、E. 由三角形外角定理,有:∠PBD=∠BPA+∠BAP、∠PCE=∠CPA+∠CAP, ∴∠PBD+∠PCE=(∠BPA+∠CPA)+(∠BAP+∠CAP)=∠BPC+∠BAC, 而∠PBD=∠PBC、∠PCE=∠PCB,∴∠PBD+∠PCE=∠PBC+∠PCB, ∴∠PBC+∠PCB=∠BPC+∠BAC,∴∠PBC+∠PCB+∠BPC=2∠BPC+∠BAC, ∴180°=2∠BPC+∠BAC,∴∠BPC=90°-(1/2)∠BAC.
蔺娟15861378940：
如图,在Rt△ABC中,∠BAC=90°,∠ACB=45°,点D是AB中点,AF⊥CD于点H,交BC于点F,BE∥AC交AF的延长线于点E,给出下列结论:①∠BAE=∠... -
10700余歪
：[选项] A. ①②⑤ B. ②④⑤ C. ①②④ D. ①②③
蔺娟15861378940：
如图,AE平分∠BAC,CE平分∠ACD,且∠1+∠2=90° 试说明CD⊥AB -
10700余歪
：[答案] 应该是证明CD//AB. 因为 AE平分∠BAC 所以 ∠2=∠BAE 因为 CE平分∠ACD 所以 ∠1=∠ECD 因为 ∠1+∠2=90° 所以 ∠ECD+∠BAE=∠1+∠2=90° 所以 ∠BAC+∠ACD =∠ECD+∠BAE+∠1+∠2 =90°+90°=180° 所以 CD//AB 见下图:
蔺娟15861378940：
如图,已知AB是⊙O的直径,P为⊙O外一点,且OP∥BC,∠P=∠BAC.求证:PA为⊙O的切线. -
10700余歪
：[答案] 证明:∵AB是⊙O的直径, ∴∠B+∠BAC=90°, ∵OP∥BC, ∴∠B=∠AOP, ∴∠POA+∠BAC=90°, ∴∠POA+∠P=90°, ∴∠OAP=180°-90°=90°, ∴OA⊥AP ∴PA为⊙O的切线.
蔺娟15861378940：
如图1,AB=AE,AC=AD,∠BAE=∠CAD=90°.(1)证明:EC=BD;(2)证明:EC⊥BD;(3)如图2,连接ED,若 -
10700余歪
： 解答:证明:(1)∵∠BAE=∠CAD=90°,∴∠BAE+∠BAC=∠CAD+∠BAC,∴∠CAE=∠BAD,在△BAD和△EAC中,,∴△BAD≌△EAC(SAS),∴EC=BD;(2)∵△BAD≌△EAC,∴∠D=∠ACE,∵∠DAC=90°,∴∠D+∠AMD=90°,∵∠AMD=∠...
蔺娟15861378940：
如图,△ABC中,∠ABC、∠EAC的角平分线PA、PB交于点P,下列结论:①PC平分∠ACF;②∠ABC+∠APC=180°; -
10700余歪
： 解:如图,过点P作PM⊥AB,PN⊥BC,PD⊥AC,垂足分别为M、N、D,①∵PB平分∠ABC,PA平分∠EAC,∴PM=PN,PM=PD,∴PM=PN=PD,∴点P在∠ACF的角平分线上(到角的两边距离相等的点在角的平分线上),故本小题正确;②∵PM⊥...
蔺娟15861378940：
已知:如图,在△ABC,△ADE中,∠BAC=∠DAE=90°,AB=AC,AD=AE,点C,D,E三点在同一条直线上,连接BD,BE.以下四个结论:①BD=CE;②∠ACE+∠... -
10700余歪
：[答案] ①∵∠BAC=∠DAE=90°,∴∠BAC+∠CAD=∠DAE+∠CAD,即∠BAD=∠CAE,∵在△BAD和△CAE中,AB=AC∠BAD=∠CAEAD=AE,∴△BAD≌△CAE(SAS),∴BD=CE,本选项正确;②∵△ABC为等腰直角三角形,∴∠ABC=∠ACB=45°...
