Let $ABCD$ be a cyclic quadrilateral, $K$ and $N$ be the midpoints of the diagonals and $P$ and $Q$ be points of intersection of the extensions of the opposite sides. Prove that $\angle PKQ + \angle PNQ = 180$. ($7$ points) .
Problem
Source: Tournament of Towns Fall 2015 Senior A-level
Tags: geometry, cyclic quadrilateral
den_thewhitelion
23.02.2017 11:48
Let $R=BD\cap AC ,OKRP=cyclic$ $O$=circumcenter of$(ABC)$ $S$=midpoint of$(PQ)$ so $S\in KN$ Now observe that the statement is equivalent with $PQ$=the common tangent of $(PNK),(QKN)$ It is well known that $R$=orthocenter of $\triangle{OPQ}$ and that the circles with diameter (PQ) and (OR) are orthogonal so $SQ^2=SP^2=SK*SN$ So $SP$ is tangent to $(PKN)$ and $(QKN)$
sunken rock
23.02.2017 20:53
Sketch: The bisectors of the angles $P$ and $Q$ are perpendicular and $PK,PN$ are isogonal lines of $\hat P$, while $QP,QN$ are isogonal lines of $\hat Q$. Best regards, sunken rock
ninjalearner
23.02.2017 22:05
Why are the bisectors of angle P and Q are perpendicular?
sunken rock
23.02.2017 22:11
ninjalearner wrote: Why are the bisectors of angle P and Q are perpendicular? Angle calculation! Best regards, sunken rock
tarzanjunior
28.07.2017 16:09
WLOG, let $AB<CD$ and $BC<AD$.
$\triangle PBD \sim \triangle PAC$ & $\triangle QBD \sim \triangle QCA$
$\implies \angle PKA = \angle PNB$ & $\angle QKC= \angle QNB$
$\implies \angle PKA + \angle QKC = \angle PNB + \angle QNB$
$\implies 180 - \angle PKQ = \angle PNQ$
Hence we get the desired result.
62861
03.11.2018 23:47
This problem is a corollary of SL 2009 G4. By SL 2009 G4, $\overline{PQ}$ is tangent to both $(PKN)$ and $(QKN)$, and the desired statement follows. In fact this is actually equivalent to 2009 G4: we know line $KN$ bisects $\overline{PQ}$ (Gauss line), so assuming the conclusion $K$ and $N$ are inverses in $(PQ)$, and the desired tangency follows.