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Is it given that CG perpendicular to ED? If not, where is point G on ED? Is it prove that angle GDF = angle OCF?

2010-05-22 07:33:54 補充：
Q1.
The intersection of the 2 diagonals of a square is a right angle, this is the property of a square, so angle DOC is a right angle.
For triangle FOC and triangle FGD
Angle FOC = angle FGD = 90 degree ( property of square and given)
Angle OFC = angle GFD ( vertically opposite angles)
therefore angle OCF = angle GDF ( angle sum of triangle).
(Note: Reason why I ask if it is given that CG perpendicular to ED is because this makes AE = EO useless).
Q2.
Angle ACD = 90 degree ( given)
Since opposite sides of rhombus or parallelogram in general are parallel, so angle BFC = 90 degree ( alternate angles BF//CD).
For triangle ABF and triangle CBF
Angle BFA = angle BFC = 90 degree ( proved)
BF = BF ( common)
angle BAF = angle BCF = 60 degree ( since triangle ABC is a equilateral triangle).
Therefore triangle ABF congruent triangle CBF ( AAS)
so AF = FC.

Because AC and BD are(對邊)
So angleBDC=BDA=45
angleDBA=DBC=ACB=ACD=CAD=CAB=45
BDC+ADC+DOC=180(angle sum of triangle)
DOC=90
DOC=90=DGC
So GOCD are concyclic(converse of angles in the same segment)
So GDF=OCF(angle in same segment)

2010-05-17 09:58:47 補充：
第2個條link睇唔到