Since $\angle ABD= \angle AED$ (A, E, B, D are concyclics)and $\angle ACD = \angle AFD$ (A, C, F, D are concyclics) we get that the triangles $ABC$ and $AEF$ are similar, from this is easy to see that $\angle EAB = \angle NAM$. Also, $\frac{AE}{AN} = \frac{AB}{AM}$ (from the similarity), then the triangles $AEB$ and $ANM$ are similar, hence $\angle ANM= \angle AEB = 90^\circ$. $Tipe$

shobber wrote: Let $ABC$ be a triangle and $D$ the foot of the altitude from $A$. Let $E$ and $F$ lie on a line passing through $D$ such that $AE$ is perpendicular to $BE$, $AF$ is perpendicular to $CF$, and $E$ and $F$ are different from $D$. Let $M$ and $N$ be the midpoints of the segments $BC$ and $EF$, respectively. Prove that $AN$ is perpendicular to $NM$. Let $k_1$ be the circle with diameter AB; this circle passes through the points A and B, through the point D (because < ADB = 90°) and the point E (because of < AEB = 90°, since $AE\perp BE$). Similarly, denote by $k_2$ the circle with diameter AC, and show that this circle passes through the points A, C, D and F. Now, we have - two circles $k_1$ and $k_2$ intersecting at two points A and D; - a line through the point D meeting the circle $k_1$ at the point B (apart from D), and the circle $k_2$ at the point C (apart from D); - a line through the point D meeting the circle $k_1$ at the point E (apart from D), and the circle $k_2$ at the point F (apart from D); - the midpoints M and N of the segments BC and EF. Hence, according to the problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=50619 , the triangles ABE, ACF and AMN are similar. Thus, < ANM = < AEB. But $AE\perp BE$ yields < AEB = 90°, and thus < ANM = 90°, so that $AN\perp NM$, and we are done. darij

As Tipe has found before $BAC$ is similar to $EAF$,. Then take $A$ as fixed by the angle rotation we can say that $BAE$ is similar to $CAF$ and $\angle BAE\cong \angle CAF$ $(i)$. Since $M, N$ are midpoints of $BC, EF$, respectively. So it is easy to see by rotation that $BAM$ is similar to $EAN$ then $\angle BAE\cong \angle MAN$ (Becouse it is angle rotation). Now look also we can say by angle rotation similarity that $FAN$ is similar to $CAM$, if we join $M$ with $N$, and $C$ with $F$ then we will get such similarity $MAN$ to $CAF$ $(ii)$ (It is theorem you can see in Geometry Revisited). From $(ii)$ $\angle CFA\cong \angle MNA$. Since $\angle CFA= 90$ so also $\angle MNA=90$. So the problem is solved. Abdurashid

Here is another easier solution: As Tipe has found before $BAC$ is similar to $EAF$,. Then take $A$ as fixed by the angle rotation we can say that $BAE$ is similar to $CAF$ and $\angle BAE\cong \angle CAF$ $(i)$. Since $M, N$ are midpoints of $BC, EF$, respectively. So it is easy to see that $BAM$ is similar to $EAN$ then $<DNA=<ENA=<BMA$, so quadrilateral $AMDN$ is cycilic, so $<MNA= <MDA$ (becouse on the same arc) since $<MDA=90$ then also $<MNA=90$. So the problem is solved. Abdurashid

$ABC$ and $AEF$ are simmilar to each other. then: $\angle AEB=\angle AFC=\angle ANM=90$ [Mod edit] In a bit more detail: clearly $\triangle ABC\sim\triangle AEF$. So $A$ is the centre of spiral similarity where $B\to E$, $C\to F$. Hence it is also the centre of spiral similarity where $B\to C$, $E\to F$. Stop this latter spiral similarity halfway through, or more generally $\frac{BM}{BC}=\frac{EN}{EF}$ through (i.e. $M,N$ don't necessarily have have to be the midpoints, just as long as those ratios are equal), to see that $\triangle ABC\sim\triangle AEF\sim AMN$, then $\angle ANM=\angle AEB=90^{\circ}$.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1998Problem4 Vo Duc Dien

Take $P, Q$ so that $PADB, AQCD$ are rectangles. Let $N$ be the midpoint of $PQ$. Then $PD$ is a diameter of the circumcircle of $ABC$, so $PX$ is perpendicular to $XY$. Similarly, $QY$ is perpendicular to $XY$. $N$ is the midpoint of $PQ$ and $M'$ the midpoint of $XY$, so $NM$ is parallel to $PX$ and hence perpendicular to $XY$. $NADM'$ is a rectangle, so $ND$ is a diameter of its circumcircle and $M$ must lie on the circumcircle. But $AM'$ is also a diameter, so $\angle AMM' = 90^{\circ}$. $\Box$

The problem is equivalent to this one: Given two circles secant at $A, D$, and $E, F$ the intersections of a line passing through $D$ with the circles, then find the locus of the point $N$, midpoint of $EF$. The locus is the circle of diameter $AM$, $M$ being the midpoint of the line segment $BC$, drawn through $D$, perpendicular to $AD$, $B, C$ lying on the two circles. Best regards, sunken rock

Dear Mathlinkers, the problem is a nice application of the mid-circle theorem. Sincerely Jean-Louis

Note that $A, B, D, E$ and $A, C, D, F$ are cyclic implies $ \triangle ABC\sim\triangle AEF $. Then, A is the center of the spiral similarity taking $B$ to $E$ and $C$ to $F$. Averaging Principle implies it takes M to N, so $ 90= \angle{ABE}= \angle{AMN}$, as required. $ \Box$

We use turning homegenity AEDB is cyclic so <ABE=<ADE. and ADCF is cyclic so <ADE= <ACF. Absolutely the triangle ABE is as same as the triangle ACF. And The vertex A is the homegenity center E goes to F. and B goes to C and we know if X and Y be on BC and EF such that BX÷ XC = EY ÷YF We have AY is altitude XY. So M and N have the situation of X and Y

Dear Mathlinkers, a reference for the mid-circle theorem http://jl.ayme.pagesperso-orange.fr/Docs/midcircle%20theorem.pdf Sincerely Jean-Louis

Solved With awesomemaths *Note we solved with a slightly different configuration, with E being to the right of ABC and F being inside* Claim - AFE is similar to ACB Proof We know $ADBE$ is a cyclic quadrilateral since opposite angles are supplementary, so $\angle{ABD} = \angle{ABD}$. We can also find that $\angle{ACB} = \angle{AFE}$ by the cyclic quadrilateral $ACDF$. By $AA$ similarity, we get the triangles are similar. Note that $\triangle{ACB}$ is sent to $\triangle{ABE}$ via the spiral similarity about A. Because of the gliding principle, we are done $\blacksquare$

very nice problem! Claim:- $\triangle{AEF} \sim \triangle{ABC}$ Pf:- we notice that points $A,B,D,E$ are concylic so we have $\angle{BAD}=\angle{BED}=90-\angle{ABC}$ and since $\angle{AEB}=90^{\circ}$ we get $\angle{AEF}=\angle{ABC}$ also since points $A,D,C,F$ are concylic we have $\angle{ACB}=\angle{ACD}=\angle{AFE}$ hence $\triangle{AEF} \sim \triangle{ABC}$ $\blacksquare$ now from Claim we have $\triangle{AEN} \sim \triangle{ABM}$ hence we get $\angle{AMD}=\angle{AND}$ as these two angles are subtended by same chord $AD$ we get points $A,D,M,N$ to be concyclic and hence we get $\angle{ANM}=180^{\circ}-\angle{ADM}=90^{\circ}$ and hence $AN \perp MN$ $\blacksquare$

Fairly straightforward spiral similarity exercise. A spiral similarity at $A$ sends $\overline{FE} \to \overline{BC}$ and thus $\overline{FN} \to \overline{BM}$. Thus $ANDM$ is cyclic and $\angle ANM = \angle ADM = 90^\circ$.

a direct application of spiral similarity. $FE \to CB \implies FN \to CM \implies ANMD $ is cyclic and we are done.

First observe trivially that $AEDB$ and $AFCD$ are cyclic. Now observe that $\triangle ABE \sim \triangle ACF$, as $\measuredangle FCA = \measuredangle FDA = \measuredangle EDA = \measuredangle EBA$. Now we make a claim which finishes this problem: Claim: $\triangle ABC \sim \triangle AEF$. Proof: \[\measuredangle AFE = \measuredangle AFD = \measuredangle ACD. \square \] Hence we now have that \[\measuredangle AMD = \measuredangle ANE = \measuredangle AND, \]so that $ADMN$ is cyclic and indeed we have $AN \perp MN$. $\blacksquare$

Remark

allaith.sh wrote: a direct application of spiral similarity. $FE \to CB \implies FN \to CM \implies ANMD $ is cyclic and we are done. very nice solution you are a genius

Let me add a coordinate bash solution . Toss the figure on the coordinate plane. Set $A \equiv (0,b), B \equiv (-1,0), C \equiv (a,0), D \equiv (0,0), M \equiv \left(\dfrac{a-1}{2},0\right)$. Let equation of line passing through the points $D,E,F$ be $y=mx$. As $\Delta ADB$ and $\Delta ADC$ are right-angled triangles, circumcenter is the midpoint of hypotenuse. Let $O_{1}$ and $O_{2}$ be the circumcenters of $\Delta ADB$ and $\Delta ADC$. Then, $$ O_{1} \equiv \left(\dfrac{-1}{2},\dfrac{b}{2}\right)$$$$ O_{2} \equiv \left(\dfrac{a}{2},\dfrac{b}{2}\right)$$Let the circumcircles of $\Delta ADB$ and $\Delta ADC$ be denoted by $\omega_{1},\omega_{2}$ respectively. Then, $$ \omega_{1} \colon x^{2}+y^{2}+x-by=0$$$$ \omega_{2} \colon x^{2}+y^{2}-ax-by=0$$Also $E$ lies on $\omega_{1}$ while $F$ lies on $\omega_{2}$ because $\angle AEB=\angle ADB=90^{\circ}$ and $\angle AFC+\angle ADC=180^{\circ}$ which implies $(A,E,D,B)$ and $(A,F,C,D)$ are pairs of concylic points. To find the coordinates of $E$ we need to find the intersection point of $y=mx$ and $x^{2}+y^{2}+x-by=0$ other than $(0,0)$. Thus, $$ x^{2}+m^{2}x^{2}+x-mbx=0$$$$ \implies x+m^{2}x+1-mb=0 \implies x=\dfrac{bm-1}{m^{2}+1}$$Thus $E \equiv \left(\dfrac{bm-1}{m^{2}+1},\dfrac{bm^{2}-m}{m^{2}+1}\right)$. Similar calculations for $F$ also yields, $F \equiv \left(\dfrac{bm+a}{m^{2}+1},\dfrac{bm^{2}+ma}{m^{2}+1}\right)$. Thus the coordinates of $N$ are given by: $$\boxed{N \equiv \left(\dfrac{bm+\dfrac{a-1}{2}}{m^{2}+1},\dfrac{bm^{2}+\dfrac{m(a-1)}{2}}{m^{2}+1}\right)}$$Now we need to show that $m_{AN}.m_{NM}=-1$ where $m_{\overleftrightarrow{l}}$ denotes slope of $\overleftrightarrow{l}$. $$m_{AN}=\dfrac{\dfrac{m(a-1)}{2}-b}{bm+\dfrac{a-1}{2}}=\dfrac{ma-m-2b}{2bm+a-1}$$$$m_{NM}=\dfrac{bm^{2}+\dfrac{m(a-1)}{2}}{bm-\dfrac{m^{2}(a-1)}{2}}=\dfrac{bm+\dfrac{(a-1)}{2}}{b-\dfrac{m(a-1)}{2}}=\dfrac{2bm+a-1}{2b+m-ma}$$Hence $m_{AN}.m_{NM}=-1 \implies AN \perp NM$ as desired. $\blacksquare$

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