Geometry : 002

Problem : Geometry
ABCD is a rectangle. I is the midpoint of CD. BI meets AC at M. Show that the line DM passes through the midpoint of BC. E is a point outside the rectangle such that AE = BE and ∠AEB = 90^o. If BE = BC = x, show that EM bisects ∠AMB. Find the area of AEBM in terms of x.
 
Answer
x²(1/2 + (√2)/3)
 
Solution
The diagonals of the rectangle bisect each other, so AC is a median of the triangle BCD, BI is another median, so M is the centroid. Hence DM is the third median and so passes through the midpoint of BC.

AB = x√2, so CI = x/√2. Hence BI = x√(3/2) and BM = (2/3)BI = x√(2/3). Now AC = x√3, so CM = (2/3)AC/2 = x/√3. Hence BC² = BM² + CM², so ∠AMB = 90^o. So M and E lie on the circle diameter AB. Hence ∠EMA = ∠EBA = 45^o, so EM bisects ∠AMB.
area AEB = x²/2, area AMB = BM·AM/2 = x√(2/3)x/√3 = x²(√2)/3. Hence area AEBM = x²(1/2 + (√2)/3)