maths問題 (急)

Q1.
Let AB = a, BC = c and AC = b. And angle C = x.
By sine rule, a/sinx = b/sin2x = b/2sinxcosx. Therefore, cosx = b/2a...............(1)
By cosine rule,
a^2 = b^2 + c^2 - 2bccosx = b^2 + c^2 - 2bc(b/2a) = b^2 + c^2 - b^2c/a.
a^3 = ab^2 + ac^2 - b^2c.
a^3 - ac^2 = ab^2 - b^2c
a(a^2 - c^2) = (a - c)b^2
a(a + c)(a -c) = (a-c)b^2
a(a + c) = b^2
b^2 = a^2 + ac. That is
AC^2 = AB^2 + AB x BC.

2008-10-09 04:28:20 補充：
Q2. Let DH, DK perpendicular to AC and BC respectively. Let AE = b , FB = c, ED = x, DF = y.
Let angle EDH = angle FDK = t. EH = ysint, DH = ycost, DK = xcost, KF = xsint. By mid-point theorem, HD = CK = KB, that is HD = KF + FB, so ycost = xsint + c or c = ycost - xsint.......(1)

2008-10-09 04:33:42 補充：
Similarly, DK = AH = AE - EH. That is xcost = b - ysint or b = xcost + ysint........(2) So c^2 + b^2 = ( ycost - xsint)^2 + ( xcost + ysint)^2. After simplification, c^2 + b^2 = x^2 + y^2. But x^2 + y^2 = EF^2 by Pythagoras theorem. That is c^2 + b^2 = EF^2 = FB^2 + AE^2.

2008-10-09 04:38:31 補充：
Correction: It should be ED = y and DF = x. They are reversed, sorry.