ABF is a right - angled triangle. In triangle ABF, pt C, D and E are on line BF and pt C, D and E are produced to A to make AC, AD and AE. angle B = 90 degrees and BC=CD=DE=EF. Prove 7(AC)square + 3(AF) square = 7(AD) square + 3(AE) square.
hope you can understand the meaning of my question coz i don't know how to put the pic on the net. IT is very urgent ......hope you can help me to solve it as soon as possible ~~~
救命ar ~畢氏定理證明
2008-05-18 9:31 am
回答 (2)
2008-05-18 5:44 pm
✔ 最佳答案
Let AB=a and BC=CD=DE=EF=k. By Pythagoras'thm:AC^2 = a^2 + k^2
AD^2 = a^2 + (2k)^2 = a^2 + 4k^2
AE^2 = a^2 + (3k)^2 = a^2 + 9k^2
AF^2 = a^2 + (4k)^2 = a^2 + 16k^2
Therefore,
7AC^2 + 3AF^2 = 7(a^2 + k^2) + 3(a^2 + 16k^2) = 10a^2 + 55k^2. and
7AD^2 + 3AE^2 = 7(a^2 + 4k^2) + 3(a^2 + 9k^2) = 10a^2 + 55k^2.
Therefore, 7AC^2 + 3AF^2 = 7AD^2 + 3AE^2.
2008-05-18 9:47 am
7(AC)square + 3(AF) square = 7(AD) square + 3(AE) square
係咪即係7*(AC)^2??
2008-05-18 01:57:08 補充：
(AB)^2 + (BC)^2 = (AC)^2
(AB)^2 + (BD)^2 = (AD)^2
(AB)^2 + (BE)^2 = (AE)^2
(AB)^2 + (BF)^2 = (AF)^2
2008-05-18 01:57:17 補充：
L.H.S.
= 7(AC)^2 + 3(AF)^2
= 10(AB)^2 + 7(BC)^2 + 3(BF)^2
= 10(AB)^2 + 7(BC)^2 + 3(4BC)^2
= 10(AB)^2 + 56(BC)^2
R.H.S
= 7(AD)^2 + 3(AE)^2
= 10(AB)^2 + 7(BD)^2 + 3(BE)^2
= 10(AB)^2 + 7(2BC)^2 + 3(3BC)^2
= 10(AB)^2 + 28(BC)^2 + 27(BE)^2
= 10(AB)^2 + 56(BC)^2
L.H.S.= R.H.S
2008-05-18 02:00:18 補充：
Sorry,
R.H.S
= 7(AD)^2 + 3(AE)^2
= 10(AB)^2 + 7(BD)^2 + 3(BE)^2
= 10(AB)^2 + 7(2BC)^2 + 3(3BC)^2
= 10(AB)^2 + 28(BC)^2 + 27(BC)^2
= 10(AB)^2 + 56(BC)^2
係咪即係7*(AC)^2??
2008-05-18 01:57:08 補充：
(AB)^2 + (BC)^2 = (AC)^2
(AB)^2 + (BD)^2 = (AD)^2
(AB)^2 + (BE)^2 = (AE)^2
(AB)^2 + (BF)^2 = (AF)^2
2008-05-18 01:57:17 補充：
L.H.S.
= 7(AC)^2 + 3(AF)^2
= 10(AB)^2 + 7(BC)^2 + 3(BF)^2
= 10(AB)^2 + 7(BC)^2 + 3(4BC)^2
= 10(AB)^2 + 56(BC)^2
R.H.S
= 7(AD)^2 + 3(AE)^2
= 10(AB)^2 + 7(BD)^2 + 3(BE)^2
= 10(AB)^2 + 7(2BC)^2 + 3(3BC)^2
= 10(AB)^2 + 28(BC)^2 + 27(BE)^2
= 10(AB)^2 + 56(BC)^2
L.H.S.= R.H.S
2008-05-18 02:00:18 補充：
Sorry,
R.H.S
= 7(AD)^2 + 3(AE)^2
= 10(AB)^2 + 7(BD)^2 + 3(BE)^2
= 10(AB)^2 + 7(2BC)^2 + 3(3BC)^2
= 10(AB)^2 + 28(BC)^2 + 27(BC)^2
= 10(AB)^2 + 56(BC)^2
收錄日期: 2021-04-15 15:19:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080518000051KK00243