Construct segment EG. Points A, E, and G lie
on a circle with center F.
Therefore,
<AGE = ½ <AFE = 30°
Using the sine rule gives,
|AE| |GE|
----------- = ----------- (1)
sin(<AGE) sin(<GAE)
|CE| |GE|
----------- = ----------- (2)
sin(<CGE) sin(<GCE)
|CE| |ED|
----------- = ----------- (3)
sin(<CDE) sin(<ECD)
|AE| |ED|
----------- = ----------- (4)
sin(<ADE) sin(<EAD)
Letting x = <EDC and y =<EGC gives
|AE| |GE|
--------- = ----------- (1')
sin(30) sin(90-y)
|CE| |GE|
-------- = --------- (2')
sin(y) sin(60)
|CE| |ED|
-------- = --------- (3')
sin(x) sin(60)
|AE| |ED|
-------- = ------------ (4')
sin(x) sin(60-2x)
Eliminating lengths from the primed eqs., gives
sin(2y) = sin(60-2x)
Either y = 30-x or y = 60+x.
<FGA + 30 = <FGA + <AGE = <FGE = <FEG = <EGC = y
If y = 30-x, then <FGA = -x. A contradiction. Thus,
y = 60+x.
Therefore,
<DAC = <EAD = 60-2x = 2(30-x) = 2(90-y)
= 2 <GAE = 2 <GAC.
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