4.如圖，以 AB 為直徑畫圓，圓周上取兩點C 、D 在 AB 異側， 已知 AB = 25， AC = 15， AD = 7 ，求CD =__________。?
回答 (1)
2019-08-23 1:32 am
AB為直徑故△ADB及△ACB皆為直角三角形，得 DB=√(25²-7²)=24，BC=√(25²-15²)=20。設AB交CD於E，則△DEB~△AEC，故AE:DE=CE:BE
=AC:DB =15:24=5:8。又△AED~△CEB，故AE:CE=DE:BE=AD:BC=7:20。
綜上得 AE/DE + BE/DE = 5/8 + 20/7
(AE+BE)/DE = 195/56
25/DE = 195/56
DE = 280/39
再者(CE:BE) : (DE:BE) = (5:8) : (7:20)
CE/DE = 25/14
(CE+DE)/DE = 39/14
CD = DE(39/14)
CD = (280/39)(39/14) = 20
=AC:DB =15:24=5:8。又△AED~△CEB，故AE:CE=DE:BE=AD:BC=7:20。
綜上得 AE/DE + BE/DE = 5/8 + 20/7
(AE+BE)/DE = 195/56
25/DE = 195/56
DE = 280/39
再者(CE:BE) : (DE:BE) = (5:8) : (7:20)
CE/DE = 25/14
(CE+DE)/DE = 39/14
CD = DE(39/14)
CD = (280/39)(39/14) = 20
收錄日期: 2021-04-11 23:00:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20190820031141AAhVMWi