Let O be the circumcenter of cyclic quadrilateral
ABCD and r the circumradius.
Part I: The maltitudes are concurrent.
Let P, Q, R, and S be the midpoints of sides
AB, BC, CD, and DA respectively. We will use
the following theorem (without proof):
A convex quadrilateral is a parallelogram
if and only if its diagonals bisect each
other.
Quadrilateral PQRS is a parallelogram since
PQ || AC || SR and PS || BD || QR. Let
G = AC ∩ BD and T the reflection of point
O about point G.
PORT is a parallelogram since |GP| = |GR| and
|GO| = |GT|. PT is perpendicular to side CD
since PT || OR and OR ⊥ CD. Therefore,
the maltitude from midpoint P to side CD
passes through point T. A similar argument
shows that the maltitude from midpoint R to
side AB passes through point T.
A similar argument, using quadrilateral QOST,
shows that maltitudes from midpoints Q and S
to sides DA and BC respectively pass through
point T.
Part II: T = Z ⇔ AC ⊥ BD.
Since ABCD is not a rectangle, at least one
of the diagonals does not pass through the
circumcenter O. WOLOG we will assume AC.
We can apply an x-y coordinate system such
that O=(0,0), A=(a,p), B=(b,q), C=(-a,p) and
D=(d,s) with
r2 = a2 + p2 = b2 + q2 = d2 + s2 and
-r ≤ s < p < q ≤ r.
From Part I we have the vector equation:
OT = OP + OR = (OA+OB)/2 + (OC+OD)/2
= (OA+OB+OC+OD)/2.
Which gives us the coordinates of point
T=( [b+d]/2 , [2p+q+s]/2 ).
The coordinates of points A, B, C, and D give
us the coordinates of point
Z=( [(p-q)(b-d)+b(q-s)]/[q-s] , p ).
Therefore,
AC ⊥ BD ⇒ d = b ⇒ s = -q ⇒
T = Z = ( b , p )
T = Z ⇒
[(p-q)(b-d)+b(q-s)]/[q-s] = [b+d]/2 (1)
and
[2p+q+s]/2 = p (2)
(2) ⇒ q+s = 0 (3)
(1) ⇒ (b-d)(2p-[q+s]) = 0 (4)
(3)&(4) ⇒ (b-d)p = 0 (5)
(5) ⇒ d = b ⇒ AC ⊥ BD.
QED
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