<s>[INST] Show that $AD=2AP$ if and only if $BP \perp CP$. Let the incircle of $\triangle ABC$ contact $AB,AC,BC$ at $D,E,F$ respectively, and intersect $AF$ at a second point $P$. Show that $AD=2AP$ if and only if $BP \perp CP$.
 [/INST] Here is a not-so-pretty partial solution.  I shall prove that, if $AD=2\cdot AP$, then $BP\perp CP$.  This is a trigonometric solution, so it is quite boring.

Without loss of generality, suppose that $\angle B\geq \angle C$.  Write $a:=BC$, $b:=CA$, $c:=AB$, and $s:=\dfrac{a+b+c}{2}$.  Note that $$AP\cdot AF=AD^2\,.$$
Thus,
$$PF=AF-AP=\frac{AF^2-AD^2}{AF}\,.$$
Let $M$ be the midpoint of $BC$.  Then,
$$PM^2=PF^2+FM^2+2\cdot PF\cdot FM\cdot\cos(\phi)\,,\tag{*}$$
where $\phi:=\angle AFB$.
We now find 
$$AF^2=AB^2+BF^2-2\cdot AB\cdot BF\cdot\cos(\angle ABC)=c^2+(s-b)^2-2c(s-b)\left(\frac{c^2+a^2-b^2}{2ca}\right)\,.$$
Thus,
$$AF^2=\frac{(s-a)}{2a}\left(a^2+(b+c)a-(b-c)^2\right)\,.\tag{#}$$
Therefore,
$$\cos(\phi)=\frac{AF^2+BF^2-AB^2}{2\cdot AF\cdot BF}=\frac{(b-c)(s-a)}{a\cdot AF}\,.$$
Note that $FM=\dfrac{b-c}{2}$. 
From (*), we get
$$PM^2=\left(\frac{AF^2-AD^2}{AF}\right)^2+\left(\frac{b-c}{2}\right)^2+\left(\frac{AF^2-AD^2}{AF}\right)(b-c)\left(\frac{(b-c)(s-a)}{a\cdot AF}\right)\,.$$
Due to (#), we get 
$$AF^2-AD^2=\frac{s-a}{2a}\left(2a^2-(b-c)^2\right)\,.$$
If $AD=2\cdot AP$, then $AF=2\cdot AD=2(s-a)=b+c-a$, since $AD=s-a$.
Using (#), we see that
$$(b+c-a)^2=AF^2=\frac{(b+c-a)}{4a}\left(a^2+(b+c)a-2(b-c)^2\right)\,,$$
so
$$0=a^2+(b+c)a-2(b-c)^2-4a(b+c-a)=5a^2-3(b+c)a-2(b-c)^2\,.$$  From (*), we have
$$PM^2=\left(\frac{3}{2}AD\right)^2+\left(\frac{b-c}{2}\right)^2+\frac{3}{4}(b-c)^2\frac{(s-a)}{a}\,.$$
Using $AD=s-a$, we conclude that
$$PM^2-\left(\frac{a}{2}\right)^2=\frac{\big(a-3(b+c)\big)\left(5a^2-3a(b+c)-2(b-c)^2\right)}{16a}=0\,.$$
Consequently, $PM=\dfrac{a}{2}=BM=CM$.  That is, the circle with diameter $M$ passes through $P$, whence $\angle BPC=\dfrac{\pi}{2}$.
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