Let $H$ be an arbitrary point on the altitude $CP$ of the acute triangle $ABC$. The lines $AH$ and $BH$ intersect $BC$ and $AC$ in $M$ and $N$, respectively. (a) Prove that $\angle NPC =\angle MPC$. (b) Let $O$ be the common point of $MN$ and $CP$. An arbitrary line through $O$ meets the sides of quadrilateral $CNHM$ in $D$ and $E$. Prove that $\angle EPC =\angle DPC$.
Problem
Source: Bulgarian MO 2003: P2
Tags: geometry unsolved, geometry
23.05.2014 11:26
The (a) part follows directly from http://www.artofproblemsolving.com/Forum/viewtopic.php?p=344654&sid=53d5bb224ccd1bab91318fa9e7037441#p344654
23.05.2014 16:03
For Part a, Note that in $\triangle ABC$, cevians $AM, BN, CP$ are concurrent. Let $MN$ intersect $AB$ at $Q$. Then $(A,B;P,Q)=-1$. Now consider the pencil $C(A,B,P,Q)$. $CA,CB,CP,CQ$ intersect the line $QN$ at $N,M,O,Q$. So $(N,M;O,Q)=-1$. But we also have $\angle OPQ=90^{\circ}$. So from the pencil $P(N,M,O,Q)$ we have that $\angle NPO=\angle MPO$ i.e. $\angle NPC=\angle MPC$. For part b, Let the arbitrary line be $\ell$. Wlog let $D$ be on the segment $CN$. Also let $\ell$ intersect $AQ$ at $F$. Now, we know that $(N,M;O,Q)=-1$. So consider the pencil $A(N,M,O,Q)$. This pencil intersects line $\el$ at $D,E,O,F$. So $(D,E;O,F)=-1$. But we also have $\angle OPF=90^{\circ}$. Thus from the pencil $P(D,E,O,F)$ we have $\angle DPO=\angle EPO$ i.e. $\angle DPC =\angle EPC$
19.08.2016 05:31
the anotheway for a, through H drawn XY//BC ( X on PM and Y on PN) by thales XH=YH and PH pedentcular XY
17.09.2016 22:24
The arbitrary line through $O$ meets which sides of $CNHM$ at $D$ and $E$?
22.03.2017 08:34
a) is Blanchet's b)Let $D\in NH$ and $E\in CM$. Let $MM\cap BC=L$ and $ED\cap AB=K$.Then $C(Q,P;A,B)=(Q,O;N,M)=B(Q,O;N,M)=(K,O;E,D)=-1$. Since $OP\perp PK$ so $\angle EPC =\angle DPC$.