[S4 MATH]Circles Q3

a)
Considering CFHD,
as angle CFH = angle HDC = 90 degree (from altitudes),
it is concluded that CFHD are concyclic.

Therefore, angle FCH = angle FDH (CFHD are concyclic).

Considering triangle ACE,
as angle CEA = 90 degree,
angle ACE + angle EAC = 180 - 90 = 90
angle FDH + angle BAC = 90
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2007-03-12 01:53:46 補充：
b)Similar as what is used in the above proof in part a), by using the properties of altitudes, it can be known that CDHE are concyclic.So, angle EBH = angle EDH.Considering triangle ACE and triangle ABF,angle EAC = angle FAB (common anlge)angle CEA = angle BFA = 90 degree

2007-03-12 01:54:28 補充：
So, angle ACE = angle ABFTherefore, angle FCH =angle FDH = angle EDH = angle EBH = say, x.Consider triangle AEF with its circumcentre K,angle FKE = 2 angle FAE = 2y (assume y = angle FAE) ---- (#)Now, consider DFKE,angle EDF = angle EDH angle FDH = 2x

2007-03-12 01:54:50 補充：
angle FKE = 2y (from #) angle EDF angle FKE= 2x 2y= 2 (x y)= 2(90 degree) --- from part a= 180 degreeTherefore, it is proved that D, F, K and E are concyclic.