When Is Triangle Isosceles:
Miguel Ochoa Sanchez's Criterion

Under the conditions of the above statement, if $AC\gt AB,\;$ then $[\Delta BCE]\gt [\Delta BCF],\;$ where $[X]\;$ denotes the area of shape $X.$

The lemma is equivalent to the assertion that point $E\;$ is "above" point $F.$

The lemma has an important consequence.

Indeed, the lemma implies that

$\displaystyle\frac{1}{2}\sin A\cdot AB\cdot AE=[ABE]\lt [ACF]=\frac{1}{2}\sin A\cdot AC\cdot AF.$ Thus we have $AB\cdot AE\lt AC\cdot AF.$

Indeed, by Ceva's theorem, $\displaystyle\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1.\;$ In addition, using the fact that $AD\;$ is an angle bisector, $\displaystyle\frac{AB}{AC}=\frac{BD}{DC}.\;$ Combining the two we get,

$\displaystyle\frac{AB-AF}{AB\cdot AF}=\frac{AC-AE}{AC\cdot AE}.$

This reduces to the required identity.

The two lemmas combined give the following:

Indeed, by Lemma 2, $\displaystyle\frac{AB+AE}{AB\cdot AE}=\frac{AC+AF}{AC\cdot AF}.\;$ So, since $AB\cdot AE\lt AC\cdot AF,\;$ so too $AB+AE\lt AC+AF.$

First we prove the following

The proof is by contradiction. WLOG, assume $AC\gt AB.\;$ Then, by Lemma 2

$\displaystyle\left(\frac{1}{AB}+\frac{1}{AE}\right)^2=\left(\frac{1}{AC}+\frac{1}{AF}\right)^2,$

implying, under our assumption, $\displaystyle\frac{2}{AB\cdot AE}=\frac{2}{AC\cdot AF}.$ But this contradicts Lemma 1.

We shall assume, as before, $AC\gt AB\;$ and now also $\displaystyle \frac{1}{AB^3}+\frac{1}{AE^3}=\frac{1}{AC^3}+\frac{1}{AF^3}.\;$ By Lemma 2,

$\displaystyle\left(\frac{1}{AB}+\frac{1}{AE}\right)^3=\left(\frac{1}{AC}+\frac{1}{AF}\right)^3,$

and subsequently,

$\displaystyle \frac{3}{AB\cdot AE\cdot (AB+AE)}=\frac{3}{AC\cdot AF\cdot (AC+AF)}.$

This is false by the corollaries to the two lemmas.

This is the original proof by Miguel Ochoa Sanchez.

Denote $\displaystyle\frac{1}{AB}+\frac{1}{AE}=s,\;$ $\displaystyle\frac{1}{AB}\cdot \frac{1}{AE}=p,\;$ $\displaystyle\frac{1}{AC}+\frac{1}{AF}=q,\;$ $\displaystyle\frac{1}{AC}\cdot\frac{1}{AF}=r.\;$

In $\Delta ABE,\;$ $\displaystyle AP=\frac{2\cos\phi}{s}\;$ and in $\Delta ACF,\;$ $\displaystyle AP=\frac{2\cos\phi}{q},\;$ implying $q=s.\;$ The given condittion is equivalent to $s^3-3sp=s^3-3sr\;$ such that $r=p.\;$ From here

$\displaystyle\left\{\frac{1}{AB},\frac{1}{AE}\right\}=\left\{\frac{1}{AC},\frac{1}{AF}\right\}.$

There are two possibilities: $AB=AF,\;AC=AE\;$ or $AB=AC,\;AE=AF.\;$ The former may only hold when $P=D,\;$ in which case the condtion of the problem becomes an inconsequential tautology and needs to be excluded. The latter possibility confirms the conclusion of the problem.

Think of other exponents.

As in Proof 3, $q=s.\;$ From the given condition, $s^4-4s^2p+2p^2=s^4-4s^2r+2r^2,\;$ implying $4s^2p-2p^2=4s^2r-2r^2.\;$ But $\displaystyle 0\lt p,r\lt\frac{s^2}{4}\;$ and the quadratic $f(x)=x^2-2s^2x\;$ is injectivebon $[0,s^2]\;$ such that $r=p.\;$ The ending as in Proof 3.

The problem above has been posted by Miguel Ochoa Sanchez at the Peru Geometrico facebook page. Proof 2 is by Miguel; Proof 3 and the resoponse to the Extra Challenge is by Leo Giugiuc.