I saw this problem about 2 years ago. It was the first time I saw the circle of Apollonius, and it still amazes me. Here is the solution that I was shown, by a friend. Pf: In my diagram, E = AB \cap CD, F = AD \cap BC. Extend CA to meet EF at a point Q. Lemma: OP bisects angle AOC. 1. By Menelaus' Thm on triangle ACF with respect to points P, B, D (clearly collinear), AP/PC*CB/BF*FD/DA=1. 2. By Ceva's Thm on triangle ACF with respect to cevians AB, CD, FQ (concurrent at E), AQ/QC*CB/BF*FD/DA=1. 3. Thus from (1) and (2), PA/PC = QA/QC, which implies that P, Q lie on the Apollonius circle of A,C with the ratio k = PA/PC = QA/QC. [I am referring to the locus of all points X such that XA/XB = k, a constant.] Since P,Q lie on the line AC, PQ must be the diameter of this circle. 4. Since QOP = 90, O lies on this circle. So, OA/OC=k=PA/PC. By the Angle Bisector Thm, it follows that angle AOP = angle POC. Similarly, by extending DB to meet EF at Q', and making analogous arguments to those above, we can show that angle DOP = angle POB. It follows that angle DOA = DOP - AOP = POB - POC = angle COB.

Call the intersection of $ AC$ and $ EF$ is $ Q$. We have that $ (ACPQ) = - 1$. On the other hand, $ OP \perp OQ$ $ \Rightarrow OP$ is bisector of $ \angle AOC$. Hence $ \angle AOD = \angle BOC$

Let $EF \cap EF = X$ Then, since $(ED,EB,EP,EX)$ is a harmonic pencil, so is $(OD,OB,OP,OX)$ that is $(OD,OB,OP,OF)$ Since $\angle POF - \frac{\pi}{2}$ $\implies OP$ is the angle bisector of $\angle BOD$ Consider $\triangle BOD$ $OB,OD$ and $OE,OF$ are isogonal lines with respect to $\angle BOD$. By Isogonal Line Lemma, since $BE \cap DE = C$ and $BF \cap DE=A$, $OC,OA$ are isogonal lines with respect to $\angle BOD$ i.e. $\angle BOC = \angle AOD$

This is a nice problem Let $AC \cap EF \equiv W$, $BD \cap EF \equiv X$, $FC \cap EP \equiv Y$, $EA \cap FP \equiv Z$. It is well known that $-1 = (F, Y; B, C) = (X, P; B, D)$. Combined with $OP \perp EF$, this means $\angle DOP = \angle BOP$. Similarly, $-1 = (A, B; Z, E) = (A, C; P, W) = -1$. As above we have $\angle AOP = \angle COP$. Thus it follows that $\angle BOC = \angle AOD$.

grobber wrote: 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 $\angle BOC=\angle AOD$. My solution. Let $X$ and $Y$ be the intersections of $EF$ with the lines $AC$ and $BD$, respectively. Because the lines $DY$,$EA$, and $FC$ are concurrent at $B$, and the points $A$,$C$, and $X$ are collinear, we deduce from Ceva's and Menelau's Theorems in the triangle $FDE$ That the cross ratio $\left(F,E,X,Y \right)=-1$, that is $\left(F,E,X,Y \right)$ is harmonic bundle. Moreover, The cross ratio $\left(C,A,X,P \right)$ is also harmonic bundle, due to the pencil $B \left(F,E,X,Y \right).$ Now, we see that the pencil $E\left(C,A,X,P \right)$ is harmonic, which means that $\left(D,B,Y,P \right)$ is harmonic bundle. Therefore, the pencil $O\left(D,B,Y,P \right)$ is harmonic, and the lines $OP$ and $OY$ are perpendicular, so we deduce that, $\angle DOP = \angle POB.$ similarly, the pencil $O \left(C,A,X,P \right)$ is harmonic, and the lines $OP$ and $OX$ are perpendicular, so we deduce that, $\angle AOP =\angle POC.$ Finally, we have $$\angle BOC=\angle POC - \angle POB=\angle AOP - \angle DOP=\angle AOD.$$

Attachments:

Lemma 1:Points $A,C,B,D$ lie on a line in this order. $P$ is a point not on this line. Then any two of the followings implie the third: $1$.$(A,B;C,D)$ is harmonic. $2$.$PB$ is the angle-bisector of $\angle CPD$. $3$.$AP\perp PB$ Proof:Sine law. Let $AC\cap EF=M$ and $EP\cap AD=K$. It's well-known that $(A,D;K,F)=-1$ $\implies$ $E(A,D;K,F)=(A,C;P,M)=-1$. Since $PO\perp OQ$ applying Lemma 1: we get the desired result.

From a famous lemma about cevians $(E,R;B,C)=-1$, projecting from $F$, we get $(Q,P;A,C)=-1$ and we are done by Apollonian circle lemma as angle $QOC=90$ Note that $R,Q$=intersection of $EP, BC;CA,EF$ Btw it's the first ever China TST problem I solved

Let $AC\cap EF = Q$ and $BD\cap EF = R$. By EGMO lemma 9.11, $-1=(E,F;R,Q)\stackrel{B}{=}(A,C;P,Q)$. Thus since $\angle POQ=90^{\circ}$ EGMO lemma 9.18 yields $\overline{PO}$ bisects $\angle AOC$. Further, EGMO lemma 9.12 yields $(B,D;P,R)=-1$. Since $\angle POR = 90^{\circ}$, EGMO lemma 9.18 yields $\overline{PO}$ bisects $\angle BOD$, as desired. $\blacksquare$

Let $Q$ be where $AC$ and $EF$ meet, and $R$ be where $BD$ and $EF$ meet. Note that $(QR; EF) = -1$ due to Ceva/Menalaus Harmonic Bundles. Thus, $(QP; CA) = -1$ and similarly $(RP; DB) = -1$. Then we see that since $\angle ROP = \angle QOP = 90^{\circ}$ and so we have that $\angle AOP = \angle COP$ and $\angle BOP = \angle DOP$ by right angles and bisectors. Finally, that gives us that $\angle AOP + \angle DOP = \angle COP + \angle BOP$ so $\angle AOD = \angle COD$ and we are done.

More storage grobber wrote: 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 $\angle BOC=\angle AOD$. Solution: Let $\overline{EP} \cap \overline{BC} = X, \overline{AC} \cap \overline{EF} = Y$ and $\overline{BD} \cap \overline{EF} = Z.$ Now, by Cevians Induce Harmonic Bundles Lemma we have $$-1=(B,C;X,F) \stackrel{E}= (A,C;P,Y) \text{ and } -1=(B,C;X,F) \stackrel{E}= (B,D;P,Z)$$Since, $\angle POY = 90^{\circ} = \angle POZ$ on applying Right Angles and Bisectors Lemma we get $\angle AOP = \angle POC$ and $\angle BOP = \angle POD$ respectively. $$\implies \angle AOB = \angle DOC \implies \angle AOD = \angle BOC. \quad \square$$

We will show the stronger result, that $\overline{OP}$ bisects both $\angle AOC$ and $\angle BOD$. Observe that $$-1 = (E, \overline{FP} \cap \overline{CD}; DC) \stackrel F= (\overline{OE} \cap \overline{AC}, P, AC),$$so $\overline{OP}$ bisects $\angle AOC$. Similar calculations yield the other bisection.

Since $$-1=(C,D;\overline{FP}\cap\overline{BC},E)\stackrel{F}{=}(C,A;P,\overline{CP}\cap\overline{FE})$$and $\angle POE=90,$ we know that $\overline{OP}$ bisects $\angle COA.$ Similarly, $\overline{OP}$ bisects $\angle BOD$ and we are done. $\square$

Let $AC\cap EF=X$. Then it is well known that $(AC;PX)=-1$, since $\angle POX=90^{\circ}$ we get $OP$ bisects $\angle AOC$. By symmetry $OP$ also bisects $\angle BOD$ and the result follows. Remark: A solution is also possible using pole-polar duality with a circle centered at $O$.

Denote $AC\cap EF = X$ and $EP \cap AD = K$. From lemma 9.11 and cevians induce harmonic bundles we get that (A, D; K, F) = -1. Now we have the pencil E(A, D; K, F) from which it follows that (A, C; P, X) = -1. We have that $OP \perp EF$ $\Rightarrow$ $\angle POX = 90^{\circ}$. So we have that (A, C; P, X) = -1 and $\angle POX = 90^{\circ}$, so we can use lemma 9.18 aka bisector and right angle lemma so we now get that $\angle COP = \angle POA$. Analogously from cevians (F, E; X, Y) is harmonic, so we have the pencil B(F, E; X, Y). From the pencil B(F, E; X, Y) it follows that (D, B; Y, P) is a harmonic bundle. We have that $OP \perp EF$ $\Rightarrow$ $\angle YOP = 90^{\circ}$. Now we have that (D, B; Y, P) is a harmonic bundle and $\angle YOP = 90^{\circ}$ so from lemma 9.18 it follows that $\angle DOP = \angle BOP$. So we got that $\angle DOP = \angle BOP$ and $\angle AOP = \angle COP$ $\Rightarrow$ $\angle DOP - \angle AOP = \angle BOP - \angle COP$ $\Rightarrow$ $\angle AOD = \angle BOC$. So we got that $\angle AOD = \angle BOC$ which was what we wanted so we are ready.

Let $I=\overline{CD} \cap \overline{FP}$. Then by Ceva-Menelaus $-1= (CD; IE) \stackrel{F}= (CA; PG)$. Therefore by Right Angles and Bisectors $\angle AOP = \angle POD$ and the conclusion follows.

Apollonian and Cevalaus Lemma gives us $\angle POB = \angle POD$, as \[(B, D; P, BD \cap EF) = (B, C; FP \cap BC, E) = -1.\] Similarily we have $\angle POA = \angle POC$, so subtracting gives the desired. $\blacksquare$