intersections of opposite sides of a convex quadrilateral

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Source: Chinese TST 2002, problem 1

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#1 h Jan 9, 2004, 11:08 PM • 5 Y

Y by Adventure10, megarnie, Mango247, and 2 other users

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 $\angle BOC=\angle AOD$ .

This post has been edited 1 time. Last edited by v_Enhance, Dec 27, 2015, 4:36 PM
Reason: Fix typo

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#2 h Mar 17, 2004, 10:30 PM • 2 Y

Y by Adventure10, Mango247

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.

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#3 h Jan 30, 2009, 4:19 AM • 2 Y

Y by Adventure10, Mango247

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 $\angle AOD = \angle BOC$

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#4 h Dec 27, 2015, 4:10 PM • 2 Y

Y by Adventure10, Mango247

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. $\angle BOC = \angle AOD$

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#5 h Mar 8, 2016, 2:11 AM • 2 Y

Y by Adventure10, Mango247

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 $\angle BOC = \angle AOD$ .

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HadjBrahim-Abderrahim

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#6 h Aug 19, 2016, 12:53 AM • 1 Y

Y by Adventure10

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:

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#7 h Mar 19, 2017, 7:51 PM • 2 Y

Y by Adventure10, Mango247

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.

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#8 h Mar 21, 2020, 4:18 PM

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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

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#9 h Aug 4, 2020, 2:28 PM

Y by

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. $\blacksquare$

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#10 h Aug 23, 2020, 3:37 PM • 1 Y

Y by Mango247

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.

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#11 h Sep 21, 2020, 8:40 PM • 2 Y

Y by A-Thought-Of-God, third_one_is_jerk

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.
$\implies \angle AOB = \angle DOC \implies \angle AOD = \angle BOC. \quad \square$

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#13 h Sep 21, 2021, 2:39 PM • 1 Y

Y by centslordm

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.

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#14 h Dec 4, 2021, 5:50 AM • 1 Y

Y by centslordm

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. $\square$

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#15 h Mar 8, 2024, 8:00 PM

Y by

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 $O$ .

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#16 h May 5, 2024, 8:12 PM

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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.

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#17 h Aug 1, 2024, 11:36 PM

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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.

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#18 h Dec 11, 2024, 4:58 PM • 1 Y

Y by GeoKing

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. $\blacksquare$

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