Show that $\angle BOC=\angle AOD$.

Question

Let $E$ and $F$ be the intersections of opposite sides of a convex quadrilateral $ABCD$. The two diagonals meet at $P$. Let $O$ be the foot of the perpendicular from $P$ to $EF$. Show that $\angle BOC=\angle AOD$.

Here's the diagram:

I defined $X=OD\cap EP, Y=EP\cap FC,Z=FP\cap EB,W=FP\cap EC $ .

Now, by a known lemma , we have $(Y,X;P,E)=-1$ and by apollonius lemma , we get $PO$ bisects $\angle XOY \implies \angle XOP =\angle POY $.

Similarly, we know that $(F,P;Z,W)=-1 \implies PO$ bisects $\angle ZOW \implies \angle ZOP =\angle WOP$ .

But this angle equalities lead me no where.Can someone give some hints ? Thanks in advance !

Answer 1

Let me please briefly rephrase the problem

A triangle $\triangle ABC$ and three cevians $AD, BE, CF$ which concur at $P$ are given. Define $O:=EF\cap AD$ and let $H$ be the orthogonal projection of $O$ onto $BC$. Prove that $\angle EHA=\angle KHF$.

Let $L:=AH\cap EF$ and $K:=HP\cap EF$. We will first prove that $\angle LHO=\angle OHK$, and then that $\angle EHO=\angle OHF$. Observe that the result follows from these observations.

For the first part, notice that -- as it is well-known -- $$-1=(D,O;P,A)\stackrel{H}=(J,O; K, L)$$ Since $(J,O; K, L)$ is harmonic and $\angle OHJ=90^\circ$, one infers that, in fact, $\angle LHO=\angle OHK$. The other part can be proven similarly, since we already have $(J,O;F,E)=-1$.