A square ABCD has a side of length 3cm. A point E on CD divides CD in such a way that CE:ED= 1:2. AC and BE intersect at P. Find cos(angle EPC). ANS= 1/(root5). approach please..

Angle EPC = ABB (Vertically opposite angle)
BAP = PCE (AI angles)
Hence the two triangles are similar so EP/PB = 1:3 = CP:AP
Let EP = x, then BP = 3x. Similarly CP = y and AP = 3y
Now using AC = 3rt2
So CP = 3rt2/4 
Similarly EP = rt10/4
We can now use the cosine formula to obtain the required value.
Hope that helps.