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

Labels:
30 degrees,
angle,
cevian,
right triangle,
triple

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http://users.sch.gr/mnannos/wp-content/uploads/2015/09/988.jpg

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