Lemma: Let ABC be a triangle; $M, N\in [BC ]$. Then $\widehat{BAM}=\widehat{CAN}\Leftrightarrow \frac{AB^2}{AC^2}=\frac{MB}{MC}.\frac{NB}{NC}$ Proof: Draw $CE \| AN, BF \| AM (E\in AB, F\in AC)$ Then $\Delta AFB\sim \Delta AEC$ so $\frac{AB^2}{AC^2} =\frac{AB}{AC}.\frac{AF}{AE}=\frac{AF}{AC}.\frac{AB}{AE}=\frac{MB}{MC}.\frac{NB}{NC}$ Let $Q\in AC, \widehat{QBC}=\widehat{MBA}$ so $\frac{BC^2}{BA^2}=\frac{MA}{MC}.\frac{QA}{QC}$ but $\frac{BC^2}{BA^2}=\frac{MA}{MC}.\frac{NA}{NB}$ $\Rightarrow \frac{QA}{QC}=\frac{NA}{NB}\Rightarrow NQ \|BC\Rightarrow BNQC$ is Isosceles trapezium We have $\widehat{BPC}=\widehat{PMC}+\widehat{PCM}=\widehat{BMC}+\widehat{NBQ}=\widehat{BMC}+\widehat{MBC}=180^{0}-\widehat{ACB}$ Let I be incircle center of $\Delta ABC$ so $\widehat{BIC}=180^{0}-\widehat{ACB}=\widehat{BPC}$ Thus P is on arc AIC

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Nice tuanh208's proof with the Steiner's theorem ! Here is and a trigonometrical proof. Quote: Let $ABC$ be a triangle with $AB=AC=b$ and $BC=a$ . Let $M\in (AC)$ , $N\in (AB)$ be two mobile points such that $\boxed{a^2\cdot AM\cdot AN=b^2\cdot BN\cdot CM}\ (*)$ and denote $P\in BM\cap CN$ . Find the locus of the point $P$ . Proof 1 (trigonometric). Denote $AN=x$ , $AM=y$ . Thus, $BN=b-x$ , $CM=b-y$ . The relation $(*)$ becomes $\underline{a^2xy=b^2(b-x)(b-y)}\ (1)$ . Denote $m\left(\widehat{CBM}\right)=\alpha$ , $m\left(\widehat{BCN}\right)=\beta$ . Thus, $m\left(\widehat{ABM}\right)=B-\alpha$ , $m\left(\widehat{ACN}\right)=B-\beta$ . Apply theorem of Sinus in triangles : $\left\|\begin{array}{cccc} \triangle\ ABM\ : & \frac {y}{\sin (B-\alpha )}=\frac {b}{\sin (B+\alpha )} & \implies & y=\frac {b\cdot\sin (B-\alpha )}{\sin (B+\alpha )}\\\\ \triangle\ ACN\ : & \frac {x}{\sin (B-\beta )}=\frac {b}{\sin (B+\beta )} & \implies & x=\frac {b\cdot\sin (B-\beta )}{\sin (B+\beta )}\\\\ \triangle\ BCM\ : & \frac {b-y}{\sin\alpha }=\frac {a}{\sin (B+\alpha )} & \implies & b-y=\frac {a\cdot\sin\alpha }{\sin (B+\alpha )}\\\\ \triangle\ BCN\ : & \frac {b-x}{\sin \beta }=\frac {a}{\sin (B+\beta )} & \implies & b-x=\frac {a\cdot\sin \beta }{\sin (B+\beta )}\end{array}\right\|$ . The relation $(1)$ becomes $\frac {a^2b^2\sin (B-\alpha )\sin (B-\beta )}{\sin (B+\alpha )\sin (B+\beta )}=\frac {b^2a^2\sin \alpha \sin\beta }{\sin (B+\alpha )\sin (B+\beta )}$ $\iff$ $\sin (B-\alpha )\sin (B-\beta )=\sin\alpha\sin\beta$ $\iff$ $\cos (\alpha -\beta )-\cos [2B-(\alpha +\beta )]=\cos (\alpha -\beta )-\cos (\alpha +\beta )$ $\iff$ $\cos [2B-(\alpha +\beta )]=\cos (\alpha +\beta )$ $\iff$ $\boxed {\alpha +\beta =B}$ . Thus, $m\left(\widehat{BPC}\right)=A+B$ (constant). Construct the rombus $ABDC$ . Observe that $D$ is $A$-exincenter and $m\left(\widehat{BDC}\right)=B$ . In conclusion, the point $P$ belongs to the circle with diameter $[ID]$ (where $I$ is incenter) so that the sideline $BC$ separates $P$ , $D$.

Isn't there any solution by Menelaus Theorem?

Proof 2 (metric). Denote $AN=x$ , $AM=y$ . Thus, $BN=b-x$ , $CM=b-y$ . The relation $(*)$ becomes $\underline{a^2xy=b^2(b-x)(b-y)}\ (1)$ . Denote $m\left(\widehat{CBM}\right)=\alpha$ , $m\left(\widehat{BCN}\right)=\beta$ . Thus, $m\left(\widehat{ABM}\right)=B-\alpha$ , $m\left(\widehat{ACN}\right)=B-\beta$ . Apply an well-known area relation $[PBN]\cdot [PCM]=[PMN]\cdot [BPC]$ . Thus $PB\cdot (b-x)\cdot \sin (B-\alpha )\cdot PC\cdot (b-y)\cdot\sin (C-\beta )=$ $=\frac {a^2\sin \alpha\sin\beta}{\sin (\alpha +\beta )}\cdot PN\cdot PM\cdot \sin (\alpha +\beta )$ $\iff$ $\frac {\sin (B-\alpha )\sin (B-\beta )}{\sin\alpha\sin \beta}=\frac {a^2}{(b-x)(b-y)}\cdot\frac {PM\cdot PN}{PB\cdot PC}\ (2)$ . Apply the Menelaus' theorem to : $\left\|\begin{array}{cc} \overline {CPN}/ \triangle ABM \ : & \frac {b-y}{b}\cdot \frac {x}{b-x}\cdot \frac {PB}{PM}=1\\\\ \overline{BPM}/ \triangle ACN\ : & \frac {b-x}{b}\cdot\frac {y}{b-y}\cdot\frac {PC}{PN}=1\end{array}\right\|$ $\bigodot\implies$ $\frac {xy}{b^2}\cdot\frac {PB\cdot PC}{PM\cdot PN}=1\ (3)$ . Therefore, $\left\|\begin{array}{cc} (1) & a^2xy=b^2(b-x)(b-y)\\\\ (2) & \frac {\sin (B-\alpha )\sin (B-\beta )}{\sin\alpha\sin \beta}=\frac {a^2}{(b-x)(b-y)}\cdot\frac {PM\cdot PN}{PB\cdot PC}\\\\ (3) & 1=\frac {xy}{b^2}\cdot\frac {PB\cdot PC}{PM\cdot PN}\end{array}\right\|\ \bigodot$ $\implies$ $\sin (B-\alpha )\sin (B-\beta)=\sin \alpha\sin\beta$ $\iff$ $\cos (\alpha -\beta )-\cos [2B-(\alpha +\beta )]=\cos (\alpha -\beta )-\cos (\alpha +\beta )$ $\iff$ $\cos [2B-(\alpha +\beta )]=\cos (\alpha +\beta )$ $\iff$ $\boxed {\alpha +\beta =B}$ . Thus, $m\left(\widehat{BPC}\right)=A+B$ (constant). Construct the rombus $ABDC$ . Observe that $D$ is $A$-exincenter and $m\left(\widehat{BDC}\right)=B$ . In conclusion, the point $P$ belongs to the circle with diameter $[ID]$ (where $I$ is incenter) so that the sideline $BC$ separates $P$ , $D$.

$I$ is the incenter of $\triangle ABC.$ The inversion with pole $A$ and radius $AB=AC=L$ transforms $\odot(IBC)$ into itself. Let $Q,M',N'$ be the inverses of $P,M,N.$ $Q \equiv \odot(ABM') \cap \odot(ACN').$ Then we have $\frac{BN'}{BN}=\frac{L}{AN} \ , \ \frac{CM'}{CM}=\frac{L}{AM} \Longrightarrow BN' \cdot CM'=L^2 \cdot \frac{BN \cdot CM}{AN \cdot AM}$ By the hypotesis, we have then $BN' \cdot CM'=BC^2 \Longrightarrow \frac{BN'}{BC}=\frac{BC}{CM'}$ Thus, $\triangle N'BC \sim \triangle BCM'$ by SAS $\Longrightarrow$ $\angle BN'C=\angle CBM'=\angle AQC$ and $\angle BCN'= \angle CM'B=\angle AQB.$ Hence, $\angle BQC=\angle BN'C+\angle BCN'=\angle ABC$ $\pmod\pi$ $\Longrightarrow$ $Q \in \odot(IBC).$ If circles $\odot(ABM')$ and $\odot(ACN')$ meet on $\odot(IBC),$ their inverse lines $BM,CN$ meet on the double circle $\odot(IBC),$ i.e. $P \in \odot(IBC).$

Quote: Let $ABC$ be a triangle with $AB=AC=b$ and $BC=a$ . Let $M\in (AC)$ , $N\in (AB)$ be two mobile points such that $\boxed{a^2\cdot AM\cdot AN=b^2\cdot BN\cdot CM}\ (*)$ and denote $P\in BM\cap CN$ . Find the locus of the point $P$ . Proof 3 (metric). Denote $AN=x$ , $AM=y$ . Thus, $BN=b-x$ , $CM=b-y$ . The relation $(*)$ becomes $\boxed{a^2xy=b^2(b-x)(b-y)}\ (1)$ . Denote $m\left(\widehat{CBM}\right)=\alpha$ , $m\left(\widehat{BCN}\right)=\beta$ . Thus, $m\left(\widehat{ABM}\right)=B-\alpha$ , $m\left(\widehat{ACN}\right)=B-\beta$ . Apply an well-known relation to assess a ratio : $\left\|\begin{array}{c} \frac {x}{b-x}=\frac {NA}{NB}=\frac {CA}{CB}\cdot\frac {\sin\widehat{NCA}}{\sin\widehat{NCB}}=\frac ba\cdot\frac {\sin (B-\beta )}{\sin \beta}\\\\ \frac {y}{b-y}=\frac {MA}{MB}=\frac {BA}{BC}\cdot\frac {\sin\widehat{MBA}}{\sin\widehat{MBC}}=\frac ba\cdot\frac {\sin (B-\alpha )}{\sin \alpha}\end{array}\right\|$ $\bigodot\ \stackrel{(1)}{\implies}$ $\sin(B-\alpha )\sin (B-\beta )=\sin \alpha\sin\beta$ $\iff$ $\cos (\alpha -\beta )-\cos [2B-(\alpha +\beta )]=$ $\cos (\alpha -\beta )-\cos (\alpha +\beta )$ $\iff$ $\cos [2B-(\alpha +\beta )]=$ $\cos (\alpha +\beta )$ $\iff$ $\boxed {\alpha +\beta =B}$ . Thus, $m\left(\widehat{BPC}\right)=A+B$ (constant). Construct the rombus $ABDC$ . Observe that $D$ is $A$-exincenter $I_a$ and $m\left(\widehat{BDC}\right)=B$ . In conclusion, the point $P$ belongs to the circle with diameter $[II_a]$ (where $I$ is incenter) so that the sideline $BC$ separates $P$ , $I_a$.

Very nice proofs above! Assuming Steiner formula is unknown, let's see another way of construction of the points $M, N$. Take a random point $M$ on $(AC)$ and $\{B, Q\}\in BM \cap \odot (ABC)$, then draw $CR||AQ$, $R$ on $\odot (ABC)$. We claim that $\{N\}\equiv AB\cap CR$ and $\{P\} \equiv BM \cap CN$ are our required points, i.e. $N$ verifies the given relation. As $\widehat{CBQ}=\widehat{CAQ}=\widehat{ACR}$ and $\widehat{ABQ}=\widehat{ACQ}$, we get $\widehat{QCR}=\widehat{ABC}\;(\;1\;)$ and, with $\widehat{QPC}=\widehat{AQB}=\widehat{ACB}\;(\;2\;)$, hence $\triangle QPC$ is isosceles, ans so is $\triangle RBP$, both of them being similar to $\triangle ABC \;(\;*\;)$. $ARCQ$ is isosceles trapezoid and, $PQ=CQ$ makes $ARPQ$ parallelogram, i.e. $AR||PQ$, consequently $\frac{BN}{AN}=\frac{BP}{PQ}$ and $\frac{CM}{AM}=\frac{CP}{PR}$ or, multiplying side by side the above two equalities: $\frac{{BN}{AN}\cdot \frac{CM}{AM}=\frac{BP}{PR}\cdot \frac{CP}{PQ}$; with$(*)$, each fraction from right side of the last equality is$\frac{BC}{AB}$, so our claim has been proved. From$(2)$we have$\widehat{QPC}=\angle B$, i.e.$\widehat{BPC}=\widehat{BIC}$and$P$lies on the arc$BIC$of$\odot (BIC)$, done. Best regards, sunken rock

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Virgil Nicula wrote: $\sin (B-\alpha )\sin (B-\beta)=\sin \alpha\sin\beta$ $\iff$ $\cos (\alpha -\beta )-\cos [2B-(\alpha +\beta )]=\cos (\alpha -\beta )-\cos (\alpha +\beta )$ Why is it true?

$\left\|\begin{array}{c} \cos (x+y)=\cos x\cos y-\sin x\sin y\\\\ \cos (x-y)=\cos x\cos y+\sin x\sin y\end{array}\right\|$ $\begin{array}{ccc} \ + & \implies & 2\cos x\cos y=\cos (x+y)+\cos (x-y)\ .\\\\ \ - & \implies & \boxed {2\sin x\sin y=\cos (x-y)-\sin (x+y)}\ .\end{array}$

Can establish somebody directly or at least using geogebra what is the geometrical locus of the interior point $P$ w.r.t. $\triangle ABC$ so that $\boxed{\frac {AM\cdot AN}{BN\cdot CM}=\frac {bc}{a^2}}$ , where $M\in AC\cap BP$ , $N\in AB\cap CP$ ?!

Virgil Nicula wrote: $\left\|\begin{array}{c} \cos (x+y)=\cos x\cos y-\sin x\sin y\\\\ \cos (x-y)=\cos x\cos y+\sin x\sin y\end{array}\right\|$ $\begin{array}{ccc} \ + & \implies & 2\cos x\cos y=\cos (x+y)+\cos (x-y)\ .\\\\ \ - & \implies & \boxed {2\sin x\sin y=\cos (x-y)-\sin (x+y)}\ .\end{array}$ Thanks very much! It was an easy equation! I don't know why i couldn't realize it!

Virgil Nicula wrote: Can establish somebody directly or at least using geogebra what is the geometrical locus of the interior point $P$ w.r.t. $\triangle ABC$ so that $\frac {AM\cdot AN}{BN\cdot CM}=\frac {bc}{a^2},$ where $M\in AC\cap BP,$ $N\in AB\cap CP.$ If $(x:y:z)$ are the barycentric coordinates of $P$ with respect to $\triangle ABC,$ then locus of $P$ is the conic $\mathcal{K} \equiv bcx^2-a^2yz=0.$ $\mathcal{K}$ passes through the incenter $I,$ A-excenter $I_a$ and is tangent to $AB,AC$ through $B,C.$ If $AB=AC$ (the proposed problem), then $\mathcal{K}$ becomes the circumcircle of $\triangle IBC.$

Thank you, Luis! It's nice result. Maybe somebody presents a draw. I'm learning just now geogebra. Thanks.

Virgil Nicula wrote: Maybe somebody presents a draw. I'm learning just now geogebra. Thanks. I also use Geogebra, since it's quite easy to manage. Note that Geogebra can only draw conic sections passing through 5 points. In this case, we need to draw $\mathcal{K}$ given 4 points $B,C,I,I_a$ and the tangent line $AC$ through $C.$ We can define a homology $\mathcal{U}$ with center $C$ mapping $\mathcal{K}$ into an arbitrary circle $\omega$ tangent to $AC$ through $C,$ in order to find as many points on $\mathcal{K}$ as we want. The following sketch shows the construction of a fith point $N$ on $\mathcal{K}.$

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