好望角234 2022-1-11 23:29
[attach]12977627[/attach]
top11 2022-1-12 12:13
[quote]原帖由 [i]好望角234[/i] 於 2022-1-11 11:29 PM 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=544429742&ptid=30377410][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]
12977627 [/quote]
(a)
∠DBA = ∠CBD (common)
∠DCB = 90° (given)
∠ADB = 90° (∠ in semi-circle)
so ∠DCB = ∠ADB.
Therefore, ΔDBA ~ ΔCBD (AA)
Thus, DB/CB = BA/BD (corr. sides, ~Δs)
That is, BD² = BC × BA.
(b)
∠ABE = ∠EBC (common)
∠BAE = ∠BEC (∠s in alt. segment)
Therefore, ΔABE ~ ΔEBC (AA)
Thus, AB/EB = BE/BC (corr. sides, ~Δs)
That is, BE² = AB × BC.
From (a), BD² = BC × BA = AB × BC.
Thus, BD² = BE² and then BD = BE.