F.2 maths.(15 marks)

在Δabc中,ab=bc. 角abc的角平分線與ac相交於d.由d作一直線與bc相交於e,使ad=de.若de//ab,求角dce
sol
設角dce=x
since ab=ac so 角abc=x
since de//ab so 角dec=x
since ad=de so 角dce=角edc so 角edc=x
so x+x+x=180度
x=60度
角dce=60度

Triangle BAD and triangle BDC are congruent (SAS), so triangle ABC is an isos. triangle, so angle BAD = angle BCD.................(1)
and AD = DC, since AD = DE, therefore, DC = DE, so triangle EDC is also an isos. triangle with angle DEC = angle BCD..........(2)
Since AD//DE, so angle BAD = angle EDC ( corr. angles AD//DE)......(3)
From (1), (2) and (3), we get
angle EDC = angle BCD = angle DEC, therefore, triangle EDC is an equilateral triangle, so angle DCE = 60 degree.

題目不明.........
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