The hexagon PQRSTU is such that:
(1) All its vertices lie on a circle, and:
(2) PQ = RS = TU, and:
(3) the diagonals PS, QT, RU meet at a point.
N is the intersection of PS and RT
Find:
RN/NT

(PR/RT)^{2}
All
angles subtended at the circumference by the three equal arcs
will be equal:
Let
angles PTQ = PRQ = RTS = RPS = TPU = TRU = a
Similarly for angles subtended by arc QR: QPR
= QTR = b
and for angles subtended by arc ST: SPT = SRT = c
and for angles subtended by arc UP: URP = UTP = d
and for angles subtended by arc PR: PSR
= PTR = a + b
and for angles subtended by arc PT: PST
= PRT = a + d
With these angles marked, and the intersection Of PS, QT and RT
denoted by O, it is easy to see that triangles ORS, SOT and RPT
are similar since they each have the angles {a+b, a+c, a+d}.
Thus OR/OS
= OS/ST = PR/RT (1)
Also triangles RNO and TNS are similar since they each have an
angle a, and the angles RNO and TNS are equal (vert. opp.).
Thus RN/NT
= OR/ST
=
OR/OS x OS/ST
=
(PR/RT)^{2} using (1)
Therefore (RN/NT)/(PR/RT)^{2}
= 1

Posted by Harry
on 20160508 15:31:37 