maths 問題 1題

為簡單起見，假設ΔABC代表埋面積。

AE:AB=2:5 => AE:EB=2:3
AG:AD=1:3 => AG:GD=1:2

ΔAEG：ΔABD=AE×AG：AB×AD=(2×1)：(5×3)=2：15
1：ΔABD=2：15，ΔABD=15/2

ΔABD：ΔABC=BA×BD：BA×BC=1：4
15/2：ΔABC=1：4，ΔABC=30

AF：FC=ΔAEG：ΔCEG
=1/3 ΔAED：|1/3 ΔCED-2/3 ΔCEA|=ΔAED：|ΔAED-2ΔCED|
=2/5 ΔABD：|3/5 ΔCAD-2*2/5 ΔCBA|=2ΔABD：|3ΔCAD-4ΔCBA|
=2*1/4 ΔABC：|3*3/4 ΔCAB-4ΔCBA|
=2：|9-16|=2：7

於是AF：AC=2：9

Why ΔAEG：ΔCEG = 1/3 ΔAED：|1/3 ΔCED-2/3 ΔCEA|?

Hint on pic03.
(1) The 6 triangles in the picture are all similar triangles.
(2) Sum of area of the 3 triangles with base x, y and z is equal to the sum of area of the 3 triangles with base p , q and r because triangle XYZ and PQR are identical.
(3) Let their individual area be X, Y, Z, P, Q and R. Using the principle of similar figure, Y = X(y/x)^2, Z = X(z/x)^2, P = X(p/x)^2, Q= X(q/x)^2 and R = X(r/x)^2.
Since X + Y + Z = P + Q + R, therefore, x^2 + y^2 + z^2 = p^2 + q^2 + r^2.

Sorry, I can only answer part (a). I'm still thinking how to solve part (b).

Area of ∆AEG = (1/2) * (length of AE) * (length of AG) * (angle EAG) = 1
Area of ∆ABD = (1/2) * (length of AB) * (length of AD) * (angle BAD)
= (1/2) * (length of AB) * (length of AD) * (angle EAG)
= (1/2) * 5/2 (length of AE) * 3(length of AG) * (angle EAG)
= (5/2)*3 [ (1/2) * (length of AE) * (length of AG) * (angle EAG) ]
= (15/2) * 1 = 15/2 = 7.5

However,
Area of ∆ABD = (1/2) * (length of AB) * (length of BD) * (angle ABD) = 7.5
Area of ∆ABC = (1/2) * (length of AB) * (length of BC) * (angle ABC)
= (1/2) * (length of AB) * (length of BC) * (angle ABD)
= (1/2) * (length of AB) * 4(length of BD) * (angle ABD)
= 4 [ (1/2) * (length of AB) * (length of BD) * (angle ABD) ]
= 4 * 7.5
= 30