We need only prove $BF^{2}=FH\cdot FA$, or $BF^{2}-FA^{2}=AH\cdot FA$(*). Put $CA=CB=x,BA=y$. Then $AH=AM\cdot \cos \angle{HAM}=\frac{2x}{3}\cdot \frac{AF^{2}+y^{2}-\frac{x^{2}}{4}}{2AF\cdot y}$. Now, I see $AF^{2}=\frac{x^{2}+y^{2}}{2}-\frac{x^{2}}{4}$ and $BF^{2}=\frac{x^{2}}{4}$. Therefore (*) is easy too.

Another proof: Invalid URL Use a homotety with centre B and factor 2. You can easily see that the problem is equivalent to prove that $AF \perp C'K$ where $\frac{AK}{KB}= \frac{1}{2}$. Now I'll use vectors. The origin is O, the circumcentre. $(A-F)(C'-K) = (A-\frac{B+C}2)(\frac{2}{3}A+\frac{1}{3}B+C) = \ldots = (\frac{2}{3}A^{2}-\frac{1}{6}B^{2}-\frac{1}{2}C^{2})+\frac{2}{3}C(A-B) = 0$.

Let $E$ be a point on $BC$ such that $CE/BE=2$ We have $ME//AC$ so $EM/AC=EB/BC=1/3=EF/FC=DE/AC$ So $ME=ED$ Because $EM//AC$ and triangle $ABC$ is isosceles we have triangle $MEB$ is isosceles so $ME=EB=ED$ Thus $\angle DBM=90=\angle MHD$ So $M,B,D,H$ are cyclic This means $\angle BHF=\angle DMB=\angle ABC$ $q.e.d$

Thank you for your nice proof

A simpler proof would be to observe that $ \tan \angle ABC = 3 \tan \angle FAB = \tan \angle BHF$

bilarev wrote: In isosceles triangle $ ABC\ ,\ (CA = CB)$ consider the point $ M\in (AB)$ so that $ MA = 2\cdot MB$ , the midpoint $ F$ of the side $ [BC]$ and the orthogonal projection $ H$ of the point $ M$ on the line $ AF$ . Prove that $ \angle BHF = \angle ABC$ . Proof. Denote the midpoint $ L$ of the side $ [AB]$ , the centroid $ G\in AF\cap CL$ of the triangle $ ABC$ , the midpoint $ N$ of the segment $ [AM]$ and the point $ D\in AF$ for which $ DB\perp AB$ . Thus, $ LN=LM$ , $ LA=LB$ , $ AN =$ $ NM =$ $ MB=$ $ 2\cdot NL$ , $ AG = GD$ , $ GC=$ $ DB=$ $ 2\cdot GL$ , $ AC\parallel GN\parallel DM$ and the quadrilateral $ MHDB$ is cyclically. Therefore, $ \widehat {BHF}\equiv$ $ \widehat {BHD}\equiv$ $ \widehat {BMD}\equiv$ $ \widehat {BAC}\equiv$ $ \widehat {ABC}$ , i.e. $ \widehat {BHF}\equiv\widehat {ABC}$ . Remark. Observe that $ \frac {MB}{ML} = \frac {AB}{AL} = 2$ , i.e. the points $ A$ , $ M$ are harmonically conjugate w.r.t. the pair $ \{L,B\}$ . Since $ HA\perp HM$ obtain $ \widehat {MHB}\equiv\widehat {MHL}$ . Thus, $ \widehat {LHA}\equiv\widehat {BHD}\equiv\widehat {FBL}$ , i.e. the quadrilateral $ LHFB$ is cyclically. Nice problem, Bilarev ! Thank you.

It was much better (I think) to take A’-symmetrical of A about F, see that D is the centroid of A’BC, hence AD/A’D=AM/MB and MD||A’B||AC, and $ \angle BMD = \angle ABC$, then BD and A’C are perpendicular, consequently $ BD\perp AB$ and BDHM cyclic, $ \angle BHD = \angle BMD$, done. Best regards, sunken rock

If you didn't see the cyclic quads... Let $G$ be the centroid of $ABC$. It suffices to show that $HBM \sim CBG$, for then $\angle GCB = \angle MHB$ implies the result. First, $\angle BGC = 180 - (90 - \angle GBA) = 90 + \angle GAB = \angle HMB$. Second, let $h_C$ denote length of altitude from $C$, and let $N$ denote the midpoint of $AM$. We have \begin{align*} \frac{HM}{CG} &= \frac{3}{2}\cdot \frac{HM}{h_C} = \frac{3}{2} \cdot \frac{[AMF]}{[ABC]} \cdot \frac{AB}{AF}\\ &= \frac{AN}{AG} = \frac{GB}{MB}. \end{align*} Therefore $HBM \sim CBG$, so the problem is finished.

I think no one posted this simple solution: Let $N$ be the midpoint of $AB$ and $G$ the centroid of $\triangle ABC$. We have $\dfrac{AG}{GF} = \dfrac{AM}{MB} = 2$ so $GM \parallel BC$. Note that $\angle MHG = 90^{\circ} = \angle GNM$ so $HNMG$ is cyclic. Furthermore $\angle ANH = \angle AGM = \angle AFB$ because $HNMG$ is cyclic and because $GM \parallel BC$. Thus $HNBF$ is cyclic and $\angle BHF = \angle BNF = \angle BAC = \angle ABC$.

Using cartesian coordinates, was only the logical way to do it . . . . We want to show that $BF^2=AF.FH$. Let $A(-3,0)$, $B(3,0)$, $C(0,c)$ meaning that $M(1,0)$ and $F\left( \frac{3}{2},\frac{c}{2} \right)$ We easily get that the line defined by the points $A$ and $F$ is the following $y=\frac{c}{9}x+\frac{c}{3}$. With this we can easily calculate the line defined with the points $H$&$M$ (not sponsored), which is the following $y=-\frac{9}{c}(x+1)$. We easily get that $H\left( -\frac{c^2+27}{c^2+81},\frac{9(c^2-54)}{c(c^2+81)} \right)$. Now all that's left to do is to see if $BF^2=AF.FH$. Which is easily seen when we plug in what we got.

see the figure;