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 $BP \perp CP$.
 A: Proof
Denote the intersection point of $DE$ and $BC$ as $G$. Then $$(AB,AC|AF,AG)=(BC|FG)=-1,\tag1$$which shows that $AF$ is the polar line of $G$ with respect to the incircle. Hence $GP$ is another tangent to the incircle, and $GP=GF.$ Moreover, notice that $DE$ is also the the polar line of $A$ with respect to the incircle. Hence $$(PF|KA)=-1.\tag2$$
Assume that $AD=2AP$. Notice that $AD^2=AP\cdot AF.$ Thus, by $(1)$ and $(2)$, we may readily obtain that $$AD:AP:PK:KF=10:5:3:12.\tag 3$$
By Menelaus theorem, we obtain $$\frac{AK}{KF}\cdot\frac{FG}{GB}\cdot\frac{BD}{DA}=1.$$
Thus $$\frac{PG}{GB}=\frac{FG}{GB}=\frac{DA}{BD}\cdot\frac{KF}{AK}=\frac{PF}{BF}\cdot\frac{DA}{AK} \cdot \frac{KF}{PF}=\frac{PF}{BF}\cdot\frac{10}{8} \cdot \frac{12}{15}=\frac{PF}{BF}.\tag 4$$
This implies that $PB$ bisects $\angle GPF$. By the properties of the harmonic point series, we may claim that $\angle BPC=90^o.$ (In fact, we may also prove that $PC$ bisects the exterior angle of $\angle GPF$, since $PG:PF=BG:BF=CG:CF$)
Till now, we have already proved that if $AD=2AP$ then $BP \perp CP.$ But obviously, by the almost perfectly similar reasoning process, we may also prove that if $BP \perp CP$ then $AD=2AP$. We are done!

A: In what follows we prove that if $AD=2AP$ then $BP \perp CP$. This proof can be rewritten so that it actually shows the equivalence of these conditions, but I'm too lazy to do it.

Let $M$ be the midpoint of $DF$ and let $PB$ intersect the incircle at $X$. Clearly $\triangle APD \sim \triangle ADF$, and therefore $\dfrac{PD}{DF} = \dfrac{AP}{AD} = \dfrac 12$. It follows that $PD = \dfrac 12 DF = DM$. 
Observe that $PB$ is the $P$-symmedian of triangle $PDF$. Angle chasing yields $\triangle PDM \sim \triangle PXF$ and since $PD=DM$, we have $PX=XF$.
We prove analogously that $PY=YF$, where we let $PC$ intersect the incircle of $ABC$ at $Y$.
Therefore $XY$ is the perpendicular bisector of $PF$, therefore $XY$ is a diameter of the incircle of $ABC$. Thus $\angle BPC = \angle XPY = 90^\circ$.
A: 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 $\angle BPC=\dfrac{\pi}{2}$.
A: Using classical geometry, the following demonstrates a sufficient condition for $AD=2AP$, and shows that $AD=2AP$ in turn implies $BP\perp CP$. 
Let $DEF$ be the incircle of triangle $ABC$, and suppose $\triangle ABC$ is isosceles, with $AC=BC$, and$$\frac{AC}{AB}=\frac{2}{1}$$Join $AF$ crossing the circle at $P$, and join $PB$, $PC$, $CD$, $EF$, and $EP$ extended to $G$.
Since $\angle BDC$ is right, then $D$ bisects $AB$, making $AE=BF$, and $ABFE$ an isosceles trapezoid with$$EF\parallel AB$$Moreover, since $\triangle ABF$ shares the angle at $B$ with $\triangle ABC$, and $$\frac{AB}{BF}=\frac{2}{1}$$then$$\triangle ABF\sim \triangle ABC$$ [Euclid, Elements VI, 6], and $\triangle ABF$ is isosceles.

Next, since $\angle AEG=\angle EFP$ [Elements III, 32], and alternate interior $\angle AGE=\angle FEP$, then$$\triangle AGE\sim \triangle EPF$$And it is evident that$$\triangle EPF\sim \triangle GPA$$Therefore$$\triangle AGE\sim\triangle GPA$$so that$$\angle AEG=\angle GAP=\angle ACB$$
and$$GE\parallel BC$$and hence$$\frac{AE}{AG}=\frac{2}{1}$$But $AE=AD$, and $AG=AP$. Therefore$$\frac{AD}{AP}=\frac{2}{1}$$.
It remains to show that $\angle BPC$ is right
In the figure below, with the same triangle $ABC$ and incircle $DEF$, draw $DJ$ parallel to $BC$ and join $PD$, $DF$, $FJ$, and $PB$. We do not know yet if $J$ lies on $PC$, and instead of drawing $PC$, join $PJ$, $JC$, and draw $JH$ tangent at $J$.
Since it is easily shown that $DBHJ$ is an isosceles trapezoid, and that $\angle JHB=\angle DBH=\angle KFB$, therefore $KF$ is equal and parallel to $JH$, so that$$\angle PFC=\angle JHC$$And since $AK$=$KF$, and $AP$ was shown to be half of $AD=AK$, making $AP=PK$, then with $AP=1$ we have $PF=3$ and $FC=6$, so that$$\frac{PF}{FC}=\frac{1}{2}$$And since $JH=DB=2$ and $HC=4$, then$$\frac{JH}{HC}=\frac{1}{2}$$
Therefore, $\triangle PFC\sim\triangle JHC$ [Elements VI, 6], and $PJ$, $JC$ lie in one straight line.
Now since by parallels$$\angle PCB=\angle PJD$$and by Elements III, 32$$ \angle PJD=\angle ADP$$then $\angle PCB$ and $\angle PDB$ are supplementary, and $PDBC$ is a cyclic quadrilateral. And since $\angle BDC$ is right, then $BC$ is the diameter of the circle through $P$, $D$, $B$, $C$, and therefore $\angle BPC$ is right.
