Let ABCD be a square. E is a variable point on the line BC, I is the midpoint of BE. CD and AE meet at F, IF and DE meet at M. The locus of M is the circumcircle of the square ABCD.
Bernard Gibert, Hyacinthos Message #19822
To prove that M lies on the circumcircle of ABCD is enough to prove that the quadrilateral BMCD is cyclic, and since DCB = 90 d., to prove that EMB = 90 d. If triangle EMB is right angled, then since IB = IE ==> IB = IE = IM.
I will prove that IE = IM.
I take the point E between B,C (if E is on the extension of BC, then signs in the calculations are changed).
Denote AB = BC = CD = DA := a and EC := x
IE = (a-x)/2
From the similar triangles FAD,FEC we get:
CF = ax / (a-x)
DF = a^2 / (a-x)
In the right triangle ICF we have:
IF^2 = IC^2 + CF^2 = (IE + EC)^2 + CF^2 = (a^2 + x^2) / 2(a-x)
Now, by Menelaus Theorem in the triangle ICF with transversal MED, we get:
MI/MF . DF/DC . EC/EI = 1
[IM /(IF-IM)]. DF/DC . EC/EI = 1
(IM / [[(a^2+x^2) / 2(a-x)] - IM]) . (a^2 / (a-x)]/a) . (x /[(a-x)/2]) = 1
==> IM = (a-x)/2
So IE = IM,