解析幾何證明。式子演算煩瑣，冗長。請鑒諒！

從同一點（O）出發有3條直線（OA,OB,OC），在3條直線上各取兩個點構成兩個三角形（A1B1C1和A2B2C2），兩個三角形對應邊的延長線各自相交形成3個交點（A1B1與A2B2相交於P點, A1C1與A2C2相交於Q點, B1C1與B2C2相交於R點），請證明這3個交點（P, Q,和R）一定是共線的。



證明：
一. 置O點於座標原點, OA在第一象限, OB是y正半軸, OC在第二象限
O點座標為（0，0）。
OA：y = k*x; OB: x = 0; OC: y = h*x。其中，斜率 k > 0, h 三角形A1B1C1: A1座標為（a1，k*a1）, B1座標為（0，b1）, C1座標為（c1，h*c1）,
其中，a1 > 0, b1 > 0, c1 三角形A2B2C2: A2座標為（a2，k*a2）, B2座標為（0，b2）, C2座標為（c2，h*c2）,
其中，a2 > 0, b2 > 0, c2 注意A和C，a和c，k和h 之間符號的對稱性，可節省重複的運算。

二. 求A1B1與A2B2交點P
A1B1: (y - b1) / (x - 0) = (k*a1 - b1) / (a1 - 0)
A1B1: a1*y - a1*b1 = (k*a1 - b1)*x
A1B1: (k*a1 - b1)*x - a1*y = -a1*b1
A2B2: (k*a2 - b2)*x - a2*y = -a2*b2
用行列式解一次方程組：
D = (k*a1 - b1)*(-a2) - (k*a2 - b2)*(-a1) = a2*b1 - a1*b2
Dx = (-a1*b1)*(-a2) - (-a2*b2)*(-a1) = a1*a2*b1 - a1*a2*b2
Dy = (k*a1 - b1)*(-a2*b2) - (k*a2 - b2)*(-a1*b1)
Dy = a2*b1*b2 - a1*b1*b2 + k*a1*a2*b1 - k*a1*a2*b2
x = Dx / D = (a1*a2*b1 - a1*a2*b2) / (a2*b1 - a1*b2)
y = Dy / D = (a2*b1*b2 - a1*b1*b2 + k*a1*a2*b1 - k*a1*a2*b2) / (a2*b1 - a1*b2)
因為A1B1與A2B2相交於P點, 方程組的解就是P點座標:
((a1*a2*b1 - a1*a2*b2) / (a2*b1 - a1*b2)，
(a2*b1*b2 - a1*b1*b2 + k*a1*a2*b1 - k*a1*a2*b2) / (a2*b1 - a1*b2))。
因為A1B1與A2B2不平行, (k*a1 - b1)/ (k*a2 - b2) != a1 / a2
即，(k*a1 - b1)*a2 != (k*a2 - b2)*a1
即，(k*a1 - b1)*a2 - (k*a2 - b2)*a1 != 0
即，a2*b1 - a1*b2 != 0

三. 求B1C1與B2C2交點R
同理於二.，利用符號的對稱性，在二.的結論中，將A和C，a和c，k和h 進行調換就得：
B1C1: (h*c1 - b1)*x - c1*y = -c1*b1
B2C2: (h*c2 - b2)*x - c2*y = -c2*b2
因為B1C1與B2C2相交於R點, 方程組的解就是 R點座標:
((c1*c2*b1 - c1*c2*b2) / (c2*b1 - c1*b2)，
(c2*b1*b2 - c1*b1*b2 + h*c1*c2*b1 - h*c1*c2*b2) / (c2*b1 - c1*b2))。
因為B1C1與B2C2不平行, c2*b1 - c1*b2 != 0

四. 求A1C1與A2C2交點Q
A1C1: (y - h*c1) / (x - c1) = (k*a1 - h*c1) / (a1 - c1)
A1C1: (y - h*c1)*(a1 - c1) = (k*a1 - h*c1)*(x - c1)
A1C1: (a1 - c1)*y - (h*a1*c1 - h*c1*c1) = (k*a1 - h*c1)*x - (k*a1*c1 - h*c1*c1)
A1C1: (k*a1 - h*c1)*x - (a1 - c1)*y = k*a1*c1 - h*a1*c1
A1C1: (k*a1 - h*c1)*x - (a1 - c1)*y = (k - h)*a1*c1
A2C2: (k*a2 - h*c2)*x - (a2 - c2)*y = (k - h)*a2*c2
用行列式解一次方程組：
D = (k*a1 - h*c1)*(c2 – a2) - (k*a2 - h*c2)*(c1 - a1)
D = k*a1*c2 - h*c1*c2–k*a1*a2 + h*a2*c1 - (k*a2*c1 - h*c1*c2–k*a1*a2 + h*a1*c2)
D = k*a1*c2 + h*a2*c1 - (k*a2*c1 + h*a1*c2)
D = (k - h)*(a1*c2- a2*c1)
Dx = (k - h)*a1*c1*(c2 – a2) - (k - h)*a2*c2*(c1 - a1)
Dx = (k - h)*(a1*c1*c2–a1*a2*c1 - a2*c1*c2 + a1*a2*c2)
Dy = (k*a1 - h*c1)*(k - h)*a2*c2 - (k*a2 - h*c2)*(k - h)*a1*c1
Dy = (k - h)*(k*a1*a2*c2 - h*a2*c1*c2– k*a1*a2*c1 + h*a1*c1*c2)
x = Dx / D = (a1*c1*c2–a1*a2*c1 - a2*c1*c2 + a1*a2*c2)/(a1*c2- a2*c1)
y = Dy / D = (k*a1*a2*c2 - h*a2*c1*c2– k*a1*a2*c1 + h*a1*c1*c2)/(a1*c2- a2*c1)
因為A1C1與A2C2相交於Q點, 方程組的解就是Q點座標:
((a1*c1*c2–a1*a2*c1 - a2*c1*c2 + a1*a2*c2)/(a1*c2- a2*c1)，
(k*a1*a2*c2 - h*a2*c1*c2– k*a1*a2*c1 + h*a1*c1*c2)/(a1*c2- a2*c1))。
因為A1C1與A2C2不平行, (k*a1 - h*c1) / (k*a2 - h*c2) != (c1 - a1) / (c2 – a2)
即，D = (k - h)*(a1*c2- a2*c1) != 0。
結果，k – h != 0, 並且a1*c2- a2*c1 != 0。
實際上,因為k > 0, 0 > h,所以, k – h > 0。

五.驗證P,Q和R 是否共線
令P(xp,yp),Q(xq,yq),和R(xr,yr),則 P,Q和R共線 當且僅當(yp–yq)/(xp - xq) = (yr–yq)/(xr - xq)。
根據二，三，和 四，
P點座標為:
((a1*a2*b1 - a1*a2*b2) / (a2*b1 - a1*b2)，
(a2*b1*b2 - a1*b1*b2 + k*a1*a2*b1 - k*a1*a2*b2) / (a2*b1 - a1*b2))。
R點座標為:
((c1*c2*b1 - c1*c2*b2) / (c2*b1 - c1*b2)，
(c2*b1*b2 - c1*b1*b2 + h*c1*c2*b1 - h*c1*c2*b2) / (c2*b1 - c1*b2))。
Q點座標為:
((a1*c1*c2–a1*a2*c1 - a2*c1*c2 + a1*a2*c2) / (a1*c2- a2*c1)，
(k*a1*a2*c2 - h*a2*c1*c2– k*a1*a2*c1 + h*a1*c1*c2) / (a1*c2- a2*c1))。

(yp - yq) = (a2*b1*b2 - a1*b1*b2 + k*a1*a2*b1 - k*a1*a2*b2) / (a2*b1 - a1*b2) –
(k*a1*a2*c2 - h*a2*c1*c2– k*a1*a2*c1 + h*a1*c1*c2) / (a1*c2 - a2*c1)
= ((a2*b1*b2*a1*c2 - a1*b1*b2*a1*c2 + k*a1*a2*b1*a1*c2 - k*a1*a2*b2*a1*c2) -
(a2*b1*b2*a2*c1 - a1*b1*b2*a2*c1 + k*a1*a2*b1*a2*c1 - k*a1*a2*b2*a2*c1) –
(k*a1*a2*c2*a2*b1 - h*a2*c1*c2*a2*b1– k*a1*a2*c1*a2*b1 + h*a1*c1*c2*a2*b1) +
(k*a1*a2*c2*a1*b2 - h*a2*c1*c2*a1*b2– k*a1*a2*c1*a1*b2 + h*a1*c1*c2*a1*b2)) /
(a2*b1 - a1*b2)* (a1*c2 - a2*c1)
= ((a1*a2*b1*b2*c2 - a1*a1*b1*b2*c2 + k*a1*a1*a2*b1*c2 - k*a1*a1*a2*b2*c2) -
(a2*a2*b1*b2*c1 - a1*a2*b1*b2*c1 + k*a1*a2*a2*b1*c1 - k*a1*a2*a2*b2*c1) –
(k*a1*a2*a2*b1*c2 - h*a2*a2*b1*c1*c2– k*a1*a2*a2*b1*c1 + h*a1*a2*b1*c1*c2) +
(k*a1*a1*a2*b2*c2 - h*a1*a2*b2*c1*c2– k*a1*a1*a2*b2*c1 + h*a1*a1*b2*c1*c2)) /
(a2*b1 - a1*b2)* (a1*c2 - a2*c1)
= ((a1*a2*b1*b2*c2 - a1*a1*b1*b2*c2 + k*a1*a1*a2*b1*c2 – X2) -
(a2*a2*b1*b2*c1 - a1*a2*b1*b2*c1 + X1 - k*a1*a2*a2*b2*c1) –
(k*a1*a2*a2*b1*c2 - h*a2*a2*b1*c1*c2– X1 + h*a1*a2*b1*c1*c2) +
(X2 - h*a1*a2*b2*c1*c2– k*a1*a1*a2*b2*c1 + h*a1*a1*b2*c1*c2)) /
(a2*b1 - a1*b2)* (a1*c2 - a2*c1)
= ((a1*a2*b1*b2*c1- a2*a2*b1*b2*c1 - a1*a1*b1*b2*c2 + a1*a2*b1*b2*c2) +
k*(a1*a1*a2*b1*c2 - a1*a2*a2*b1*c2–a1*a1*a2*b2*c1 + a1*a2*a2*b2*c1) -
h*(a1*a2*b1*c1*c2–a2*a2*b1*c1*c2 - a1*a1*b2*c1*c2 + a1*a2*b2*c1*c2)) /
(a2*b1 - a1*b2)* (a1*c2 - a2*c1)
= ((a1–a2)*a2*b1*b2*c1–(a1–a2)*(a1*b1*b2*c2) +
k*((a1–a2)*a1*a2*b1*c2–(a1–a2)*a1*a2*b2*c1) -
h*((a1–a2)*a2*b1*c1*c2–(a1–a2)*a1*b2*c1*c2)) /
(a2*b1 - a1*b2)* (a1*c2 - a2*c1)
= ((a1–a2)*b1*b2*(a2*c1–a1*c2)+
k*(a1–a2)*a1*a2*(b1*c2–b2*c1) -
h*(a1–a2)*c1*c2*(a2*b1–a1*b2)) /
(a2*b1 - a1*b2)*(a1*c2 - a2*c1)
= (a1–a2)*(b1*b2*(a2*c1–a1*c2)+ k*a1*a2*(b1*c2–b2*c1)- h*c1*c2*(a2*b1–a1*b2)) /
(a2*b1 - a1*b2)*(a1*c2 - a2*c1)

(xp - xq)
= (a1*a2*b1 - a1*a2*b2) / (a2*b1 - a1*b2) – (a1*c1*c2–a1*a2*c1 - a2*c1*c2 + a1*a2*c2) / (a1*c2 - a2*c1)
= ((a1*a2*b1 - a1*a2*b2)*(a1*c2 - a2*c1)–(a1*c1*c2–a1*a2*c1 - a2*c1*c2 + a1*a2*c2)*(a2*b1 - a1*b2)) /
(a2*b1 - a1*b2)*(a1*c2 - a2*c1)
= ((a1*a2*b1 - a1*a2*b2)*(a1*c2 - a2*c1)–(a1*c1*c2–a1*a2*c1 - a2*c1*c2 + a1*a2*c2)*(a2*b1 - a1*b2)) /
(a2*b1 - a1*b2)*(a1*c2 - a2*c1)
= ((a1*a2*b1*a1*c2 - a1*a2*b2*a1*c2)–(a1*a2*b1*a2*c1 - a1*a2*b2*a2*c1)–
((a1*c1*c2*a2*b1–a1*a2*c1*a2*b1 - a2*c1*c2*a2*b1 + a1*a2*c2*a2*b1) –
(a1*c1*c2*a1*b2–a1*a2*c1*a1*b2 - a2*c1*c2*a1*b2 + a1*a2*c2*a1*b2))) /
(a2*b1 - a1*b2)*(a1*c2 - a2*c1)
= ((a1*a1*a2*b1*c2–a1*a1*a2*b2*c2–a1*a2*a2*b1*c1 + a1*a2*a2*b2*c1)–
(a1*a2*b1*c1*c2–a1*a2*a2*b1*c1–a2*a2*b1*c1*c2 + a1*a2*a2*b1*c2) +
(a1*a1*b2*c1*c2–a1*a1*a2*b2*c1–a1*a2*b2*c1*c2 + a1*a1*a2*b2*c2)) /
(a2*b1 - a1*b2)*(a1*c2 - a2*c1)
= ((a1*a1*a2*b1*c2–X1– X2 + a1*a2*a2*b2*c1)–
(a1*a2*b1*c1*c2– X2–a2*a2*b1*c1*c2 + a1*a2*a2*b1*c2) +
(a1*a1*b2*c1*c2–a1*a1*a2*b2*c1– a1*a2*b2*c1*c2 + X1)) /
(a2*b1 - a1*b2)*(a1*c2 - a2*c1)
= (a1*a1*a2*b1*c2–a1*a1*a2*b2*c1–a1*a2*a2*b1*c2 + a1*a2*a2*b2*c1 +
a1*a1*b2*c1*c2–a1*a2*b1*c1*c2–a1*a2*b2*c1*c2 + a2*a2*b1*c1*c2)) /
(a2*b1 - a1*b2)*(a1*c2 - a2*c1)
= (a1*a1*a2*(b1*c2–b2*c1)–a1*a2*a2*( b1*c2–b2*c1) +
a1*c1*c2*(a1*b2–a2*b1)– a2*c1*c2*(a1*b2 - a2*b1)) /
(a2*b1–a1*b2)* (a1*c2 - a2*c1)
= ((a1*a1*a2–a1*a2*a2)*(b1*c2–b2*c1) + (a1*c1*c2– a2*c1*c2)*(a1*b2 - a2*b1)) /
(a2*b1 - a1*b2)* (a1*c2 - a2*c1)
= (a1*a2*(a1–a2)*(b1*c2–b2*c1) + c1*c2*(a1– a2)*(a1*b2 - a2*b1)) /
(a2*b1 - a1*b2)* (a1*c2 - a2*c1)
= ((a1–a2)*(a1*a2*(b1*c2–b2*c1) + c1*c2*(a1*b2 - a2*b1)) /
(a2*b1 - a1*b2)* (a1*c2 - a2*c1)

(yp–yq)/(xp - xq)
= (b1*b2*(a2*c1–a1*c2)+ k*a1*a2*(b1*c2–b2*c1)- h*c1*c2*(a2*b1–a1*b2)) /
(a1*a2*(b1*c2–b2*c1) + c1*c2*(a1*b2 - a2*b1))

利用符號的對稱性，在上式中，將a和c，k和h 進行對換就得：
(yr–yq)/(xr - xq)
= (b1*b2*(c2*a1–c1*a2)+ h*c1*c2*(b1*a2–b2*a1)- k*a1*a2*(c2*b1–c1*b2)) /
(c1*c2*(b1*a2–b2*a1) + a1*a2*(c1*b2 - c2*b1))
= (b1*b2*(a2*c1–a1*c2)+ k*a1*a2*(b1*c2–b2*c1)- h*c1*c2*(a2*b1–a1*b2)) /
(a1*a2*(b1*c2–b2*c1) + c1*c2*(a1*b2 - a2*b1))

(yp–yq)/(xp - xq) = (yr–yq)/(xr - xq)

P,Q和R共線
證明畢。

所有跟帖： 

• 哇，太牛了！ 這個一定要強頂... -與數學無關- ♂ (449 bytes) () 09/18/2010 postreply 00:47:15

• 謝！有時候發現一個正確的命題不亞於驗證這個命題是正確的。 -皆兄弟也- ♂ (22 bytes) () 09/19/2010 postreply 08:23:57

• 是。初三的時候。當時盲目激動了一下，嗬嗬~ -與數學無關- ♂ (0 bytes) () 09/19/2010 postreply 12:57:15

• 天才出少年。如果當今仍是世界獨一份，可以投稿，命名類似Pascal定理。 -皆兄弟也- ♂ (21 bytes) () 09/19/2010 postreply 13:50:36

• 啊，沒有，應該不會是獨一份的嗬嗬... -與數學無關- ♂ (647 bytes) () 09/19/2010 postreply 16:08:21

• 哈！很有啟發哎！用反證法：如果二維不共線，可構造一三維例，也不共線。 -皆兄弟也- ♂ (0 bytes) () 09/19/2010 postreply 19:37:00

• 嗯，用反證法也許敘述起來會更為容易一些。 -與數學無關- ♂ (0 bytes) () 09/20/2010 postreply 09:28:36

• 三維空間裏，用“平麵相交於直線”的公理偷了很大的懶，所以容易。 -皆兄弟也- ♂ (58 bytes) () 09/19/2010 postreply 22:13:06

• thank you, -jinjing- ♀ (0 bytes) () 09/18/2010 postreply 17:40:47

• you are right that it is easy by theory, but boring in practice. -皆兄弟也- ♂ (0 bytes) () 09/19/2010 postreply 08:30:46