Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 988
Friday, February 28, 2014
Geometry Problem 988: Right Triangle, Cevians, Angles, 30 Degrees, Triple Angle
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ReplyDeleteA slightly modified approach
ReplyDeleteMark X on AD extended and Y on AC such that EA = EX = EY
Now E is the centre of circle AXY and so < XEY = 2 X < XAY = 60
Hence Tr. XEY is equilateral and EA = EX = EY = XY
Now < ADE = 3α – (60+ α) = 2α – 60 and so < XED = (2α – 60) – (α – 30) = α – 30 = < EAX
Therefore XE^2 = XD.XA = XY^2 and so < XYD = < XAY = 30
But < YCD = 90-α = < YXD hence CDXY is concyclic and so < DCX = 30 and hence XY = XC
It follows that Y is the centre of Tr. CEY and so x = < EXY = 60/2 = 30
Sumith Peiris
Moratuwa
Sri Lanka
Sumith, congratulations for your solution. Certainly difficult to find. Best Regads, Joaquim Maia, Brazil
ReplyDeleteThank you Joaquim
DeleteBest Rgds
Sumith