d06624d299e6b4e3363e0fc705ebda3a029eaa87c208a516d983abaf1989acce 5555406|手续费: 0.006 DOGE|2,317 Satoshis/vByte 2025/01/22 16:53:41 DECKb2wXgznhgn11wxxaguTXTVuZWHcoQn 140.96879991 D73p6qMgnK2bHNFBwCkZTjS5XtDUgtzwLx 31.51 DQcY22oVU1NPKBx8xU6KbJ2AcbwkrgAF7i 41.55 DECKb2wXgznhg...
(1)过C点作CN⊥DE垂足为N,∵△ABC≌△DEC,∴AB=DE.∵S△ABC=0 2 AB•Cn=S△DCE=1 2 DE•CN,∵CF=CN,∴CP平分∠EPA.(2)如图2在4A上截取4M=4E连接CM,过C作CK⊥4A,由(1)同理可证CP平分∠EPA,∴∠EPC=∠APC.∵PM=PB PC=PC,∴△PsC≌△PEC,∴CE=CM,PE=PM.又∵CE=CB,∴CM=CB=5...
[MS-MSB]: Media Stream Broadcast (MSB) Protocol [MS-MSB]: Media Stream Broadcast (MSB) Protocol 1 Introduction 2 Messages 3 Protocol Details 4 Protocol Examples 4 Protocol Examples 4.1 General MSB Sequence 4.2 Server-Side Playlist Streaming by Using MSB 4.3 NSC File Encoding 5 Security 6 Appe...
【解析】(1)过C点作CN⊥DE垂足为N∵△ABC≡△DEC,.AB=DE.S△ABC=02AB⋅Cn=S_△DCE= 12DE·CN,CF=CN,∴.CP270*27 2)上截取4M=4E连接CM,过C作CK⊥4A,(1)√(12)+√2C 由P平分∠EPA,∠PM=PB PC=PC,∴.△PsC≡△PECCE=CM,PE=PM .又∵CE=CB∴.CM=CB=5,且CK⊥PA,∴K为BM的中...
[答案]B[解析][分析]利用勾股定理求出AB=10,利用等积法求出CN=(24)/5,从而得AN=,再证明∠NMC=∠NCM=45°,进而即可得到答案.[详解]解:∵∠ACB=90°,AC=8,BC=6∴AB=√(4e^2+BC^2)=1^2÷8^2=10^2,∵S△ABC=×AB×CN=×AC×BC∴CN=(24)/5,∵AN=,∵折叠∴AM=A'M,∠BCN=∠B'...
解：（1）过C点作CN⊥DE垂足为N，∵△ABC≌△DEC，∴AB=DE．∵S△ABC= 0 2 AB•Cn=S△DCE= 1 2 DE•CN，∵CF=CN，∴CP平分∠EPA．（2）如图2在4A上截取4M=4E连接CM，过C作CK⊥4A，由（1）同理可证CP平分∠EPA，∴∠EPC=∠APC．∵PM=PB PC=PC，∴△PsC≌△PEC，...
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【题目】已知:任意一个三角形的三条角平分线都交于一点,如图,在△ABC中,BD,CD分别平分∠ABC,∠ACB,过点D作直线分别交AB,AC于点E,F,若AE=AF,解答下列问题:(1)证明:DE=DF;(2)若∠A=60°,AB=8,BC=7,AC=5,求EF的长4EDFB℃ 答案 【解析】(1)连接ADBD,CD分别平分∠ABC,∠ACB∴AD平分∠BAC又∵...
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