(*) if ${OQ}\perp{EF}$ we have that $OQCF$ is cyclic (because, by simmetry, ${OC}\perp{AC}$). so, $\angle{OFE} = \angle{OCB} = \angle{OBC}$. ($OB = OC$) see that $EQOB$ is cyclic, because $\angle{OBE}+ \angle{EQO} = 180$. so, $\angle{OEQ} = \angle{OBC} = \angle{OFE} \Longrightarrow OE = OF \Longrightarrow QE = QF$. (**) if $QE = QF$. by menelaus, we have $\frac{EB . AC . QF}{AB . CF . QE} = 1$, but $AB = AC$ and $QE = QF$. $\Longrightarrow CF = EB$. but, $OB = OC$, so, $\triangle{EBO} \equiv \triangle{OCF}$ $\Longrightarrow OE = OF \Longrightarrow {OQ}\perp{EF}$ and the problem is solved

This solution isn't very different from the nice E.Lopes's solution. I note the reflection $E'$ of the point $E$ w.r.t. the point $B\ (BE=BE')$, i.e. the line $BO$ is the bisector of the segment $EE'$. From the Menelaus' theorem applied to the transversal $\overline {BQC}$ for the triangle $AEF$ results the relation $\frac{BE}{BA}\cdot \frac{CA}{CF}\cdot \frac{QF}{QE}=1$, i.e. $\frac{BE'}{CF}=\frac{QE}{QF}$. Thus, $QE=QF\Longleftrightarrow BE'=CF\Longleftrightarrow E'F\parallel BC\Longleftrightarrow$ $AO$ is the bisector of the segment $[E'F]\Longleftrightarrow$ the point $O$ is the circumcenter of the triangle $E'EF\Longleftrightarrow OQ\perp EF$.

I have the same solution as E.Lopes's

Here is a slightly different solution for when $ QE=QF$:

Nemion wrote: I have the same solution as E.Lopes's Did you have to revive a 2 year old thread to just say this?

Virgil Nicula wrote: This solution isn't very different from the nice E.Lopes's solution. I note the reflection $ E'$ of the point $ E$ w.r.t. the point $ B\ (BE = BE')$, i.e. the line $ BO$ is the bisector of the segment $ EE'$. From the Menelaus' theorem applied to the transversal $ \overline {BQC}$ for the triangle $ AEF$ results the relation $ \frac {BE}{BA}\cdot \frac {CA}{CF}\cdot \frac {QF}{QE} = 1$, i.e. $ \frac {BE'}{CF} = \frac {QE}{QF}$. Thus, $ QE = QF\Longleftrightarrow BE' = CF\Longleftrightarrow E'F\parallel BC\Longleftrightarrow$ $ AO$ is the bisector of the segment $ [E'F]\Longleftrightarrow$ the point $ O$ is the circumcenter of the triangle $ E'EF\Longleftrightarrow OQ\perp EF$. really nice solution dear experienced virgil i will post my solution later!

An easy extension. Let $ w$ be the circumcircle of the triangle $ ABC$ . Let $ [AA']$ be the diameter of the circle $ w$ . Consider three collinear points $ D\in (BC)$ , $ E\in AB$ , $ F\in AC$ . Prove that $ A'D\perp EF\iff\frac {A'E}{A'F} = \frac {AB}{AC}$ .

Virgil Nicula wrote: An easy extension. Let $ w$ be the circumcircle of the triangle $ ABC$ . Let $ [AO]$ be the diameter of the circle $ w$ . Consider three collinear points $ Q\in (BC)$ , $ E\in AB$ , $ F\in AC$ . Prove that $ OQ\perp EF$ if and only if $ \frac {QE}{QF} = \frac {AB}{AC}$ . nice extension but i don't think it's completly correct(we can't express $ \frac{QE}{QF}$ in terms of $ b$ and $ c$ only,I think)but i think if we change problem a little it would work . Problem.Let $ \omega$ be the circumcircle of the triangle $ ABC$ . Let $ AA'$ be the diameter of the circle $ \omega$ . Consider three collinear points $ D\in BC$ , $ E\in AC$ , $ F\in AF$ . Prove that $ A'D\perp EF$ if and only if $ \frac {A'E}{A'F} = \frac {A'C}{A'B}$ Solution.Let $ A'D\perp EF$ we have $ A'B\perp AB$ therefore $ BDA'F$ is cyclic hence $ \angle DFA'=\angle DBA'=\angle A'AC$ hence $ AEA'F$ is cyclic by similarity of Triangles $ EA'F$ and $ BA'C$ we have : $ \frac{A'F}{A'E}=\frac{A'B}{A'C}$ . now consider $ \frac{A'F}{A'E}=\frac{A'B}{A'C}$ hence $ \frac{A'F}{A'B}=\frac{A'E}{A'C}$ and $ \angle A'CE=\angle A'BF=\frac{\pi}{2}$ hence $ \angle A'FA=\angle A'EC$ hence $ AEA'F$ is cyclic. therefore $ \angle A'BC=\angle EAA'=\angle EFA'$ therefore $ BDA'F$ is cyclic then $ \angle FDA'=\frac{\pi}{2}$

For A different(own) solution, see here

Dear Mathlinkers, for the initial problem, see http://www.artofproblemsolving.com/Forum/index.php?f=47&t=330790. Sincerely Jean-Louis

For $QE=QF\Rightarrow OQ\perp EF$ i have a different approach, consider the triangle $AEF$, since the perpendiculars from $O$ to the sides $AE, EF, FA$ of these triangle are collinear, we have by the Simson lines that $AEOF$ is cyclic and then $\measuredangle{OEF}=\measuredangle{OAF}=\measuredangle{OAE}=\measuredangle{OFE}$, since $ABC$ is isosceles, then $OQ\perp EF$, qed.

In fact there is a spiral similarity with centre $O$ than sends $\triangle OBE$ to $\triangle OCF$, I feel very dim for not noticing this at first. Instead an almost equivalent solution by easy angle-chasing: WLOG $E$ is on the segment $AB$ and $F$ is on the ray $AC$ extended beyond $C$. First assume $OQ$ is perpendicular to $EF$, then $\angle OQE=\angle OBE=90^{\circ}$ then $OQEB$ is a cyclic quadrilateral (perhaps degenerate). So $\angle OBQ=\angle OEQ$. Next since $\angle OCF=90^{\circ}=\angle FQO$ so $QCOF$ is a cyclic quadrilateral which implies $\angle OQF=\angle QCO=\angle MCO=\angle MBO=\angle OEQ\implies\triangle OEF$ is isosceles. So $QE=QF$. For the next direction, use Menelaus on $AEF$ wrt $BQC$ to see $\triangle BEO$ is congruent to $\triangle OCF$. Then $OE=OF$ so $OQ$ is the median implies $OQ$ is the perp. bisector so $OQ\perp EF$.

i have a solution which is very similar to that of e.lopes

I have a solution which is not at all similar to that of e.lopes!

My solution: First suppose $OQ\perp EF$ then $BQOE$ and $OQFC$ are both cyclic and thus $\angle OEQ=\angle OBQ=\angle BAM=\angle MAC=\angle MCO=\angle OFQ$ implies $EQ=QF$ as desired. Now suppose $EQ=QF$. Observe that as $\angle EBQ+\angle QCF =180^{\circ}$ and $\angle EQB=\angle DQF$ and $EQ=QF$, this means $\Delta QBE$ and $\Delta QCF$ are both part of the ambiguous case. Hence $EB=FC$ & also $OB=OC$ So $\Delta EBO$ is congruent to $\Delta FCO$ implying $OF=OE$.

I don't know why but the problem in the IMO Compendium is a bit different and more difficult: N is an arbitrary point on the bisector of ∠BAC. P and O are points on the lines AB and AN, respectively, such that ∡ANP = 90◦ = ∡APO. Q is an arbitrary point on NP, and an arbitrary line through Q meets the lines AB and AC at E and F respectively. Prove that ∡OQE = 90◦ if and only if QE = QF .

$ \angle ABC=x,\angle OQB=y$ By Sine rule in $\Delta OQB$, $\frac{BQ}{\sin(90-y+x)}=\frac{BO}{\sin y}\implies BQ=\frac{c\tan (90-x)\cos(x-y)}{\sin y}$ By Sine Rule in $\Delta OQC$, $\frac{QC}{\sin(90-x-y)}=\frac{b\tan (90-x)}{\sin(180-y)}\implies QC=\frac{b\cot x \cos(x+y)}{\sin y}$ $\implies \frac{BQ}{CQ}=\frac{\frac{c\tan (90-x)\cos(x-y)}{\sin y}}{\frac{b\cot x \cos(x+y)}{\sin y}}$ $=\frac{c}{b}\cdot \frac{\cos(x-y)}{\cos(x+y)} =\frac{\cos(x-y)}{\cos(x+y)}$ ==================== $OQ\perp EF\implies QE=QF$ ==================== $\implies \angle QEB=y-x+90,\angle CFQ=x+y-90(\angle OQE=90)$ By Sine rule in $\Delta BEQ$ $\frac{\sin x}{QE}=\frac{\sin (y-x+90)}{BQ}$ By Sine rule in $\Delta QCF$ $\frac{\sin (180-x)}{QF}=\frac{\sin (x+y-90)}{QC}$ Dividing these two, we get, $\frac{QF}{QE}=\frac{\frac{-\cos (x+y)}{QC}}{\frac{\cos(y-x)}{BQ}}=\frac{-\cos(x+y)}{\cos (y-x)}\cdot\frac{BQ}{QC}=\frac{-\cos(x+y)}{\cos (y-x)}\cdot \frac{\cos(x-y)}{\cos(x+y)}=1$ $\implies QE=QF$. ==================== $QE=QF\implies OQ\perp EF$ ==================== Let $\angle EPQ=p$ By Sine Rule in $\Delta EBQ$, $\frac{\sin(180-(p+x))}{BQ}=\frac{\sin x}{EQ}$ By Sine Rule in $\Delta FQC$ $\frac{\sin (x-p)}{QC}=\frac{\sin(180-x)}{QF}$ $\implies \frac{\sin (p+x)}{\sin (x-p)}\cdot \frac{QC}{BQ}=\frac{FQ}{EQ}$ $\implies \frac{\sin (p+x)}{\sin (x-p)}=\frac{BQ}{CQ}=\frac{\cos(x-y)}{\cos(x+y)}$ $\implies \frac{\sin (p+x)}{\sin (x-p)}=\frac{\cos(x-y)}{\cos(x+y)}$ After expanding, it is easy to get $p=90-y\implies \angle OQE=90-y+y=90\implies OQ\perp EF$

Indeed you trigo bashed it bravely. But @elope's solution is easier and same to the one I got.

My solution: By Perpendicularity lemma we have: $OQ\perp ED$$\Longleftrightarrow$$OE^2-OF^2=QE^2-QF^2$. Since $OB\perp AB$ and $OC\perp AC$ so $OF^2=OC^2+CF^2$ and $OE^2=OB^2+BE^2$ and $OB=OC$. So we have $OE^2-OF^2=BE^2-CF^2$. From law of sine on($\angle EQB=x$,$\angle ABC=b$) $\triangle QBF$:$ BE=\frac{\sin x}{\sin b}QE$. Analogously $CF=\frac{\sin x}{\sin b}QF$. So $OE^2-OF^2=BE^2-CF^2$ so $OQ\perp ED$$\Longleftrightarrow$$OE^2-OF^2-BE^2+CF^2=0$$\Longleftrightarrow$ $(OE^2-OF^2)(\frac{sin^2(x)-sin^2(b)}{sin^2(b)}=0$ $ \Longleftrightarrow$ $QE=QF$(if $x=b$ or $x=180-b$ $EF$ is parallel to $AC$ or $AB$).

The feet of the perpendiculars from $O$ to the sides of $\triangle AEF$ are $B,Q,C$, and are collinear. By the converse of Simson, this means that $OEAF$ is cyclic. Therefore, $\angle FAO = \angle FEO$. Since \[ \angle FAO = 90^\circ-\angle ACM = 90^\circ - (180^\circ-\angle BCO - 90^\circ) = \angle BCO ,\]we know that $\angle FEO = \angle BCO$. Now, we see that $\angle FEO = \angle EFO \iff EQ=QF$. Also, \[ \angle BCO = \angle EFO \iff OFCQ \text{ cyclic } \iff \angle FGO = \angle OCF = 90^\circ.\]Therefore, $EQ=QF \iff \angle EQF = 90^\circ$.

Here's a (hopefully) different solution. First let's tackle the if $OQ \perp EF$ condition: Note $O$ is the antipode to $A \implies \angle OBA =90^{\circ} \implies OQBE$ is cyclic. Similarly $OCFQ$ is cyclic $\implies \angle OEQ = \angle OBC = \angle OCB = \angle OFQ \implies OF=OE$. As $OFE$ is isosceles and $OQ \perp EF \implies QE=QF$. Now for the if $QE=QF$ condition: Let $E'$ be the reflection of $E$ in $B$ and $F'$ the reflection of $F$ in $C$. First note $OE=OE'$, as $BE=BE'$ and $OB \perp EE'$. First, we claim $BE=CF$. To see why, let's apply sine rule on triangles $BQE$ and $CQF$: $\frac{BE}{\sin{\angle BQE}} = \frac{QE}{\sin{\angle QBE}}$ and $\frac{FC}{\sin{\angle FQC}} = \frac{QF}{\sin{\angle FCQ}}$. Note that $\angle QBE=\angle OBE +\angle QBO=90^{\circ}+\angle OAC=180^{\circ}-\angle ACB =180^{\circ}-\angle FCQ \implies \sin{\angle FCQ}=\sin{\angle QBE}$. As $\angle BQE =\angle FQC$ and $QE=QF \implies BE=CF$. Thus $AE \cdot AE'=AB^2-BE^2=AC^2-CF^2=AF \cdot AF'$, aka $AE'FF'$ is cyclic, with circumcenter $O$, as $OE=OE' \implies OE=OF$. As $Q$ is the midpoint of $EF$, we conclude $QO \perp EF$.

Proving that if $OQ \perp EF$, $QE = QF$: $BQOE$ is cyclic because $\angle EBO = \angle EQO = 90$. Therefore, $\angle QEO = \angle OBQ$. Since $OB^2 = AO^2 - AB^2$ and $OC^2 = OA^2 - AC^2 = OA^2 - AB^2$, $OC = OB$, so $\angle OBQ = QCO$. Since $\angle FQO + \angle FCO = 90 + 90 = 180$, $QOFC$ is cyclic with $\angle QCO = \angle QFO$. Putting this all together, $\angle QEO = \angle QFO$, so $EO = OF$. $EQ^2 = EO^2-QO^2$ and $QF^2 = FO^2 - QO^2 = EO^2 - QO^2$. Thus, $QE = QF$. Proving that if $QE = QF$, $OQ \perp EF$: Note that $\angle EBO = \angle FCO = 90$. Since $OB^2 = AO^2 - AB^2$ and $OC^2 = OA^2 - AC^2 = OA^2 - AB^2$, $OC = OB$. Now, if $\angle BEQ = b-a$ and $\angle BQE =a$, then $\angle BQO = 90-b \Longleftrightarrow \angle QCO = 90-b \Longleftrightarrow \angle FCQ = b$. It’s also true that $\angle BQE = \angle FQC = a$. Therefore, by the Law of Sines, we have that: $$\frac{QE}{\sin (180-b) } = \frac{BE}{\sin (a)} \Longleftrightarrow \frac{QE}{\sin (b) } = \frac{BE}{\sin (a)}$$$$\frac{QF}{\sin (b)} = \frac{FC}{\sin (a)}$$Therefore, $BE = FC$, and $\triangle BOE$ is congruent to $\triangle COF$, which implies that $EO = FO$, which in turn implies that $\triangle EQO$ and $\triangle FQO$ are congruent. This means that $OQ \perp EF$.

Firstly , if $OQ \perp{EF}$ , Clearly $\angle{OCF}=\frac{\pi}{2}$ , thus $OQCF$ is cyclic , hence $\angle{BFQ}=\angle{OFQ}=\angle{OCB}=\angle{OBC}=\angle{FBQ}$ wich implies that $QF=QB$ , and since $\angle{FBE}=\frac{\pi}{2}$ , thus Q is the circumcenter of $\triangle{BEF}$ , hence $QF=QE$. Now if $QF=QE$ , similary $Q$ is the circumcenter of $\triangle{BEF}$ , thus $\angle{BEQ}=\angle{B}=\angle{C}=\angle{QCA}$ , hence $QCAE$ is cyclic , this implies that $\angle{EQB}=\angle{A}$ , since $ABOC$ is cyclic $\angle{A}=\pi-\angle{BOC}=\angle{COF}$ , thus $\angle{CQF}=\angle{EQB}=\angle{COF}$ , hence $CQOF$ is cyclic , $\longrightarrow \angle{OQF}=\angle{OCF}=\frac{\pi}{2}$ .

To show that "if QE = QF then (OQ) is perpendicular to (EF)", you can bash it quite easily with a repere. All points have some decent coordinates so it works fine.

If Angle OQE=90, QE=QF (obviously) If Angle OQE is not 90, Make a line E'F' that supports E'F'ㅗOQ QE'=QF' If QE=QF, Square E'EF'F becomes a parallelogram, so it is false. So, QE=QF can only be true when Angle OQE=90 So, Angle OQE=90<=>QE=QF