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

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 !

• Are you using projective geometry because you are supposed to solve the problem with this tool, or was it rather your particular approach? Commented Aug 17, 2020 at 16:42
• yes, I am supposed to use projective geo .. Commented Aug 17, 2020 at 16:45
• @Shubhangi, which book are you using? I need such questions for practice. Commented Aug 17, 2020 at 16:56
• @SarGe , I am using EGMO and A beautiful Journey through Olympiad Geometry plus Evan Chen's cross ratio handout . Commented Aug 17, 2020 at 16:57

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 $$(J,O;F,E)=-1$$.