I am going to give two solutions, but first I rewrite the problem. In fact, the equation < NAC = < MAB shows that the line AN is the isogonal of the line AM wrt the angle CAB. Thus, the point N lies on the isogonal of the line AM wrt the angle CAB. Similarly, < NBC = < MBA shows that the point N lies on the isogonal of the line BM wrt the angle ABC. Hence, the point N is the point of intersection of the isogonals of the lines AM and BM wrt the angles CAB and ABC. But it is known that the isogonals of the lines AM and BM wrt the angles CAB and ABC intersect at the isogonal conjugate of the point M wrt triangle ABC. Hence, the point N is the isogonal conjugate of the point M wrt triangle ABC. Therefore, solving the problem is equivalent to proving the following theorem: Theorem 1. Let M be a point in the plane of a triangle ABC, and let N be the isogonal conjugate of the point M wrt triangle ABC. Then, $\frac{AM \cdot AN}{AB \cdot AC}+\frac{BM \cdot BN}{BC \cdot BA}+\frac{CM \cdot CN}{CA \cdot CB}= 1$. First proof of Theorem 1. Since N is the isogonal conjugate of the point M wrt triangle ABC, the line AN is the isogonal of the line AM wrt the angle CAB. Hence, < NAB = < CAM (in fact, with directed angles modulo 180°, we would have < (AB; AN) = - < (CA; AM), but we don't use directed angles here). We will use the notation $\left[P_{1}P_{2}...P_{n}\right]$ for the (non-directed) area of an arbitrary polygon $P_{1}P_{2}...P_{n}$. Let X', Y', Z' be the reflections of the point M in the lines BC, CA, AB. Since Z' is the reflection of M in the line AB, we have AZ' = AM, < BAZ' = < BAM and [AZ'B] = [AMB]. Thus, < NAZ' = < NAB + < BAZ' = < CAM + < BAM = < CAB. Since the area of a triangle equals $\frac12$ times the product of two of its sides times the sine of the angle between them, we have $\left[ABC\right]=\frac12\cdot AB\cdot AC\cdot\sin\measuredangle CAB$ and $\left[NAZ^{\prime}\right]=\frac12\cdot AZ^{\prime}\cdot AN\cdot\sin\measuredangle NAZ^{\prime}$. Hence, $\frac{\left[NAZ^{\prime}\right]}{\left[ABC\right]}=\frac{\frac12\cdot AZ^{\prime}\cdot AN\cdot\sin\measuredangle NAZ^{\prime}}{\frac12\cdot AB\cdot AC\cdot\sin\measuredangle CAB}=\frac{\frac12\cdot AM\cdot AN\cdot\sin\measuredangle CAB}{\frac12\cdot AB\cdot AC\cdot\sin\measuredangle CAB}=\frac{AM\cdot AN}{AB\cdot AC}$. Similarly, $\frac{\left[NBZ^{\prime}\right]}{\left[ABC\right]}=\frac{BM\cdot BN}{BC\cdot BA}$. Thus, $\frac{AM\cdot AN}{AB\cdot AC}+\frac{BM\cdot BN}{BC\cdot BA}=\frac{\left[NAZ^{\prime}\right]}{\left[ABC\right]}+\frac{\left[NBZ^{\prime}\right]}{\left[ABC\right]}=\frac{\left[NAZ^{\prime}\right]+\left[NBZ^{\prime}\right]}{\left[ABC\right]}$ $=\frac{\left[NAZ^{\prime}B\right]}{\left[ABC\right]}=\frac{\left[AZ^{\prime}B\right]+\left[ANB\right]}{\left[ABC\right]}$, what, using [AZ'B] = [AMB], becomes $\frac{AM\cdot AN}{AB\cdot AC}+\frac{BM\cdot BN}{BC\cdot BA}=\frac{\left[AMB\right]+\left[ANB\right]}{\left[ABC\right]}$. Similarly, $\frac{BM\cdot BN}{BC\cdot BA}+\frac{CM\cdot CN}{CA\cdot CB}=\frac{\left[BMC\right]+\left[BNC\right]}{\left[ABC\right]}$; $\frac{CM\cdot CN}{CA\cdot CB}+\frac{AM\cdot AN}{AB\cdot AC}=\frac{\left[CMA\right]+\left[CNA\right]}{\left[ABC\right]}$. Summing up these three equations, we obtain $2\cdot\left(\frac{AM\cdot AN}{AB\cdot AC}+\frac{BM\cdot BN}{BC\cdot BA}+\frac{CM\cdot CN}{CA\cdot CB}\right)$ $=\frac{\left[BMC\right]+\left[BNC\right]+\left[CMA\right]+\left[CNA\right]+\left[AMB\right]+\left[ANB\right]}{\left[ABC\right]}$. Thus, $\frac{AM\cdot AN}{AB\cdot AC}+\frac{BM\cdot BN}{BC\cdot BA}+\frac{CM\cdot CN}{CA\cdot CB}$ $=\frac{\left[BMC\right]+\left[BNC\right]+\left[CMA\right]+\left[CNA\right]+\left[AMB\right]+\left[ANB\right]}{2\cdot\left[ABC\right]}$ $=\frac{\left(\left[BMC\right]+\left[CMA\right]+\left[AMB\right]\right)+\left(\left[BNC\right]+\left[CNA\right]+\left[ANB\right]\right)}{2\cdot\left[ABC\right]}$ $=\frac{\left[ABC\right]+\left[ABC\right]}{2\cdot\left[ABC\right]}=1$, and Theorem 1 is proven. Second proof of Theorem 1. First remember the easy fact that if a quadrilateral has perpendicular diagonals, then its area equals $\frac12$ of the product of the two diagonals. In the same way as in the First proof of Theorem 1, we show that the point N is the isogonal conjugate of the point M wrt triangle ABC. Hence, by a well-known property of isogonal conjugates (Theorem 6 in [1]), it follows that if X, Y, Z are the orthogonal projections of the point M on the lines BC, CA, AB, then $AN\perp YZ$, $BN\perp ZX$ and $CN\perp XY$. Hence, the quadrilaterals AYNZ, BZNX and CXNY have perpendicular diagonals, and thus, their areas are $\left[ AYNZ\right] =\frac12 \cdot AN\cdot YZ$; $\left[ BZNX\right] =\frac12 \cdot BN\cdot ZX$; $\left[ CXNY\right] =\frac12 \cdot CN\cdot XY$. Hereby, [F] denotes the area of a figure F. Since < AYM = 90° and < AZM = 90°, the points Y and Z lie on the circle with diameter AM; thus, the segment AM is a diameter of the circumcircle of the triangle AYZ. Hence, by the Extended Law of Sines, $YZ = AM \cdot \sin \measuredangle YAZ = AM \cdot \sin A$. Hence, $\left[ AYNZ\right] =\frac12 \cdot AN\cdot YZ = \frac12 \cdot AN \cdot AM \cdot \sin A$. But if S is the area of triangle ABC, then $S = \frac12 \cdot AB \cdot AC \cdot \sin A$. Thus, $\frac{AM\cdot AN}{AB\cdot AC}= \frac{\frac12 \cdot AN\cdot AM \cdot \sin A}{\frac12 \cdot AB \cdot AC \cdot \sin A}= \frac{\left[ AYNZ\right] }{S}$. Similarly, $\frac{BM\cdot BN}{BC\cdot BA}= \frac{\left[ BZNX\right] }{S}$; $\frac{CM\cdot CN}{CA\cdot CB}= \frac{\left[ CXNY\right] }{S}$. Altogether, $\frac{AM\cdot AN}{AB\cdot AC}+\frac{BM\cdot BN}{BC\cdot BA}+\frac{CM\cdot CN}{CA\cdot CB}= \frac{\left[ AYNZ\right] }{S}+\frac{\left[ BZNX\right] }{S}+\frac{\left[ CXNY\right] }{S}$ $= \frac{\left[ AYNZ\right]+\left[ BZNX\right]+\left[ CXNY\right] }{S}= \frac{\left[ABC\right]}{S}= \frac{S}{S}= 1$, what proves Theorem 1 again. References [1] Darij Grinberg, Isogonal conjugation with respect to a triangle, avaliable at http://de.geocities.com/darij_grinberg/ and http://www.mathlinks.ro/Forum/viewtopic.php?t=18472. Darij

orl wrote: Let $ M$ and $ N$ be points inside triangle $ ABC$ such that \[ \angle MAB = \angle NAC\quad \mbox{and}\quad \angle MBA = \angle NBC. \] Prove that \[ \frac {AM \cdot AN}{AB \cdot AC} + \frac {BM \cdot BN}{BA \cdot BC} + \frac {CM \cdot CN}{CA \cdot CB} = 1. \] Sorry to revive an old topic, but here is another solution:

Lemma: In $\triangle ABC$, let $a = BC$, $b = CA$, and $c = AB$, and let $P$ be a point in the interior of $\triangle ABC$ with barycentric coordinates $(p,q,r)$ (with $p$ corresponding to $A$, $q$ corresponding to $B$, and $r$ corresponding to $C$.) Let $U$ and $V$ be points such that $\angle UAV = \pi - \angle A$, $AU = rb$ and $AV = qc$. Then $AP (p+q+r) = UV = \sqrt{(br)^2 + (cq)^2 + (rq)(b^2 + c^2 - a^2)}$. Proof: Let $AP$ hit $BC$ at $D$. By Stewart's theorem, $a \cdot AD^2 + a \cdot BD \cdot DC = (AB)^2(DC) + (AC)^2 (BD)$. $BD = \frac{ar}{q+r}$ and $DC = \frac{aq}{q+r}$, so $AD^2 a + \frac{a^3 qr}{(q+r)^2} = \frac{ac^2 q + ab^2 r}{q+r}$, so $AD^2 = \frac{c^2 q + b^2 r}{q+r} - \frac{a^2 qr}{(q+r)^2} = \frac{(br)^2 + (cq)^2 + rq(b^2 + c^2 - a^2)}{(q+r)^2}$. By the law of cosines, $b^2 + c^2 - a^2 = 2 bc \cos A$. Hence, $\begin{align*} (br)^2 + (cq)^2 + (rq)(b^2 + c^2 - a^2) &= (br)^2 + (cq)^2 + 2(rq)(bc) \cos A \\ &= (br)^2 + (cq)^2 - 2(br)(cq) \cos(\pi - A),$ which by the law of cosines is equal to $UV^2$. Therefore, $AD(q+r) = UV$. From properties of barycentric coordinates, $\frac{AP}{PD} = \frac{q+r}{p}$, so $\frac{PD}{AP} = \frac{p}{q+r}$, so $\frac{AD}{AP} = 1 + \frac{PD}{AP} = \frac{p+q+r}{q+r}$, so $AP = \frac{AD(q+r)}{p+q+r} = \frac{UV}{p+q+r}$, as desired. In $\triangle ABC$, let $a = BC$, $b = CA$, and $c = AB$, and let the barycentric coordinates of $M$ be $(p,q,r)$. The trilinear coordinates of $M$ are thus $(\frac{p}{a}, \frac{q}{b}, \frac{r}{c})$. The conditions of the problem imply that $N$ is the isogonal conjugate of $M$, so the trilinear coordinates of $N$ must be $(\frac{a}{p}, \frac{b}{p}, \frac{c}{r})$. It follows that the barycentric coordinates of $N$ are $(\frac{a^2}{p}, \frac{b^2}{q}, \frac{r^2}{q})$. Let $X$ and $Y$ be points in space such that $\angle XAY = \pi - \angle A$. Pick $B_1$ and $C_1$ on rays $\overrightarrow{AX}$ and $\overrightarrow{AY}$, respectively, so that $AB_1 = rb$ and $AC_1 = qc$. Pick $B_2$ and $C_2$ on rays $\overrightarrow{AX}$ and $\overrightarrow{AY}$, respectively, so that $AB_2 = \frac{c^2}{r} b$ and $AC_2 = \frac{b^2}{q} c$. Note that $\frac{AC_2}{AB_2} = \frac{\frac{b^2}{q} c}{\frac{c^2}{r} b} = \frac{br}{cr} = \frac{AB_1}{AC_1}$, so $\triangle AB_1 C_1 \sim \triangle A C_2 B_2$. Hence, $\frac{B_1 C_1}{AB_1} = \frac{B_2 C_2}{AC_2}$, so ${(B_1 C_1)(B_2 C_2) = (B_1 C_1)^2 \frac{AC_2}{AB_1} = (B_1 C_1)^2 \frac{\frac{b^2}{q}c}{rb}} = (B_1 C_1)^2 \frac{bc}{qr}$. By our lemma, we find that \begin{align*} \frac{AM \cdot AN}{AB \cdot AC} &= \frac{(B_1 C_1)(B_2 C_2)}{(p + q + r)\left(\frac{a^2}{p} + \frac{b^2}{q} + \frac{c^2}{r}\right) \cdot bc} \\ &= \frac{(B_1 C_1)^2 \frac{bc}{qr}}{(p + q + r)\left(\frac{a^2}{p} + \frac{b^2}{q} + \frac{c^2}{r}\right) \cdot bc} \\ &= \frac{p(br)^2 + p(cq)^2 + pqr(b^2 + c^2 - a^2)}{(a^2 qr + b^2 pr + c^2 pq)(p+q+r)}. \end{align*} Hence, \begin{align*} \frac{AM \cdot AN}{AB \cdot AC} + \frac{BM \cdot BN}{BA \cdot BC} + \frac{CM \cdot CN}{CA \cdot CB} &= \sum_{cyc} \frac{p(br)^2 + p(cq)^2 + pqr(b^2 + c^2 - a^2)}{(a^2 qr + b^2 pr + c^2 pq)(p+q+r)} \\ &= \frac{1}{(a^2 qr + b^2 pr + c^2 pq)(p+q+r)} \left( \sum_{cyc} b^2 p^2 r + \sum_{cyc} c^2 p q^2 + pqr \sum_{cyc} (b^2 + c^2 - a^2) \right) \\ &= \frac{1}{(a^2 qr + b^2 pr + c^2 pq)(p+q+r)} \left( \sum_{cyc} b^2 p^2 r + \sum_{cyc} b^2 p^2 q + pqr(a^2 + b^2 + c^2) \right) \\ &= \frac{(a^2 qr + b^2 pr + c^2 pq)(p + q + r)}{(a^2 qr + b^2 pr + c^2 pq)(p+q+r)} \\ &= 1, \end{align*} as desired.

Just a small note: since $M,N$ are clearly isogonal conjugates, \[\frac{(m-a)/(b-a)}{(c-a)/(n-a)}\in\mathbb{R},\]which is thus equal to its magnitude. The result quickly follows. The identity involved here also shows by the triangle inequality that \[\sum\frac{AM\cdot AN}{AB\cdot AC}\ge1\]for arbitrary points $M,N$ in the plane, with equality (I think?) iff $M,N$ are isogonal conjugates.

This problem can be killed of without the use of isogonal conjugates(expect the fact that if three cevians are concurrent,their isogonal lines are also concurrent).Just apply sine rule in $\triangle{AMB},\triangle{BNC},\triangle{CNA},\triangle{BMC}$to get the ratios $\frac{AM}{AB},\frac{AN}{AC},\frac{BM}{AB},\frac{BN}{BC},\frac{CN}{AC},\frac{CM}{BC}$ in terms of the angles and noting that $\angle{BCM}=\angle{ACN}$ and the result will follow.

sayantanchakraborty wrote: This problem can be killed of without the use of isogonal conjugates(expect the fact that if three cevians are concurrent,their isogonal lines are also concurrent).Just apply sine rule in $\triangle{AMB},\triangle{BNC},\triangle{CNA},\triangle{BMC}$to get the ratios $\frac{AM}{AB},\frac{AN}{AC},\frac{BM}{AB},\frac{BN}{BC},\frac{CN}{AC},\frac{CM}{BC}$ in terms of the angles and noting that $\angle{BCM}=\angle{ACN}$ and the result will follow.

Could anyone show the complete steps for this solution? I don't see how the result follows once we get everything in terms of the sines those angles.

1998 G4 wrote: Let $ M$ and $ N$ be two points inside triangle $ ABC$ such that \[ \angle MAB = \angle NAC\quad \mbox{and}\quad \angle MBA = \angle NBC. \]Prove that \[ \frac {AM \cdot AN}{AB \cdot AC} + \frac {BM \cdot BN}{BA \cdot BC} + \frac {CM \cdot CN}{CA \cdot CB} = 1. \] Funny solution Note that $M$ and $N$ are isogonal conjugates in $\triangle ABC$. Lemma. Let line $AM$ meet $\odot(BMC)$ again at $M_A$. Then $\triangle AM_AB \sim \triangle ACN$. Proof. Clearly, $\angle BAM_A=\angle NAC$ and $\angle AM_AB=\angle MM_AB=\angle MCB=\angle ACN$ proving the similarity. $\blacksquare$ Thus, $\sum_{\text{cyc}} \frac{AM \cdot AN}{AB \cdot AC}=\sum_{\text{cyc}} \frac{AM}{AM_A}$. Now we apply inversion at $M$, to convert this into the equivalent problem. inverted problem wrote: Point $M$ lies in the interior of triangle $ABC$ and lines $AM, BM, CM$, meet sides $BC, CA, AB$ at $M_A, M_B, M_C$, respectively. Then show $\sum_{\text{cyc}} \frac{MM_A}{AM_A}=1$. However, this is obvious: if $M$ has barycentric coordinates $(x:y:z)$ then we are just summing $\frac{x}{x+y+z}$ and the conclusion follows. $\blacksquare$

Let $[\cdot]$ denote area. Also, let $\triangle DEF$ be the pedal triangle of $M$ with respect to $\triangle ABC.$ Claim: $\frac{AM\cdot AN}{AB\cdot AC}=\frac{[AENF]}{[ABC]}$ Proof. Notice $\overline{EF}\perp\overline{AN}$ as $$\angle NAE+\angle AEF=\angle FAM+\angle AMF=90.$$By LoS, $EF=AM\cdot\sin\angle BAC,$ and we also know $[ABC]=\tfrac{1}{2}\cdot AB\cdot AC\cdot\sin\angle BAC.$ Thus, $$2[AENF]=AN\cdot EF=AM\cdot AN\cdot\sin\angle BAC=\frac{2[ABC]\cdot AM\cdot AN}{AB\cdot AC}.$$$\blacksquare$ $$\frac{AM\cdot AN}{AB\cdot AC} + \frac{BM\cdot BN}{BA\cdot BC} + \frac{CM\cdot CN}{CA\cdot CB}=\frac{[AENF]+[BDNF]+[CDNE]}{[ABC]}=1.$$$\square$

Let ray $AN$ meet $(BNC)$ again at $A_1$, ray $BN$ meet $(CNA)$ again at $B_1$, and ray $CN$ meet $(ANB)$ again at $C_1$. Clearly, $M$ and $N$ are isogonal conjugates wrt $ABC$. Observe that $$\angle BAM = \angle NAC = \angle A_1AC$$and $$\angle ABM = \angle NBC = \angle NA_1C = \angle AA_1C$$yielding $ABM \sim AA_1C$. Thus, $$\frac{AM \cdot AN}{AB \cdot AC} = \frac{AC}{AA_1} \cdot \frac{AN}{AC} = \frac{AN}{AA_1}.$$Analogous processes give $$\frac{AM \cdot AN}{AB \cdot AC} + \frac{BM \cdot BN}{BA \cdot BC} + \frac{CM \cdot CN}{CA \cdot CB} = \frac{AN}{AA_1} + \frac{BN}{BB_1} + \frac{CN}{CC_1}$$so we're done by BAMO 2008/6, which is also EGMO 8.26. $\blacksquare$ Remarks: The sole purpose for constructing $A_1$, $B_1$, and $C_1$ is to get rid of $M$ entirely. Also, EGMO 3.18 is the inverted version of EGMO 8.26.

Let $BN$ intersect $(ANC)$ again at $P$. Note that $\angle BPC=\angle NAC=\angle BAM$ and $\angle ABM=\angle NBC$ implying that $\triangle BAM\sim \triangle BPC$. Similarly, $\triangle BCM\sim \triangle BPA$. $~$ Thus, we have \begin{align*} &~~~~\frac {AM \cdot AN}{AB \cdot AC} + \frac {BM \cdot BN}{BA \cdot BC} + \frac {CM \cdot CN}{CA \cdot CB}\\ &= \frac{AM}{AB}\cdot \frac{AN}{AC}+\frac{BM}{BA}\cdot \frac{BN}{BC}\cdot \frac{CM}{CB}\cdot \frac{CN}{CA}\\ &= \frac{PC}{PB}\cdot \frac{AN}{AC}+\frac{BC}{BP}\cdot \frac{BN}{BC}+\frac{AP}{BP}\cdot \frac{CN}{CA}\\ &= \frac{PC\cdot AN + BN\cdot AC+AP\cdot CN}{BP\cdot AC} \\ &= \frac{AC\cdot PN + AC\cdot BN}{BP\cdot AC} = 1 \end{align*}as desired.

Employ barycentric coordinates on $\triangle ABC$. Note that $M$ and $N$ are isogonal conjugates, so letting $M=(p:q:r)$ gives $N=\left(\frac{a^2}{p}:\frac{b^2}{q}:\frac{c^2}{r}\right)$. Applying distance formula and simplifying, we have $$\frac{AM\cdot AN}{AB\cdot AC} = \frac{-a^2p q r + b^2 p q r + b^2p r^2 + c^2 p q^2 + c^2 p q r}{a^2 p q r + a^2 q^2 r + a^2 q r^2 + b^2 p^2 r + b^2 p q r + b^2 p r^2 + c^2 p^2 q + c^2 p q^2 + c^2 p q r}.$$From here we can see that we will add back the $a^2pqr$ twice, and similar with the other variables, so the numerator will contain $a^2pqr+b^2pqr+c^2pqr$. Then, by symmetry, all the terms similar to $b^2pr^2$ will be counted once in our sum, so the numerator is exactly the denominator, which means that the LHS is equal to $1$, as desired.

Note that $M$ and $N$ are isogonal conjugates, so \[\frac{(m-a)(n-a)}{(b-a)(c-a)}\]is positive and real. Therefore, the LHS is equal to \begin{align*} & \sum_{\text{cyc}}\frac{(m-a)(n-a)}{(b-a)(c-a)} \\ =& -\sum_{\text{cyc}}\frac{(m-a)(n-a)(c-b)}{(b-a)(a-c)(c-b)} \\ =& -\frac{1}{(b-a)(a-c)(c-b)}\sum_{\text{cyc}}\Big(mn(c-b)+(-ac+ab)(m+n)+a^2(c-b)\Big) \\ =& -\frac{1}{(b-a)(a-c)(c-b)}\sum_{\text{cyc}}a^2(c-b) \\ =& 1\;\blacksquare \end{align*}