ELMO Shortlist 2012/G3: Geometry involved angle bisector and right angle 
Given $\triangle ABC$ with $I$ as its incenter, draw $ID$ perpendicular to $BC$ and $IP$ perpendicular to $AD$. Prove that $\angle BPD=\angle DPC$

It seems like just a normal classic geometry with incenter and some perpendicular, but still I wasn't able to solve it. The solution of this problem involving Pole/Polar or sth like that, that I dun study or know anything about it.
Here is the link to the problem and solution
I'm a highschool student who like to solve geometry and I want to solve it using just normal method without involving higher theorems and here is my approach:
Let $E,F$ be another touching point on $AB$ and $AC$ respectively.
Since $IP\perp AD$ it suggests that $A,E,I,P,F$ lie on the same circle with diameter $AI$
Let $J$ be the midpoint of $AI$ so $IJ\perp EF$
Denote $M,N$ be the point where $PC, PB$ meet $(J)$
So $\angle BPD=\angle DPC$ if and only if $AM=AN$ or $EF\parallel MN$ or $EM=FN$.
Another way around is to prove that
\begin{equation}
\frac{BD}{DC}=\frac{BP}{PC}
\end{equation}
Unfortunately, that's all about that I can do. Please help
To simplify the pic I only take the important part because if I take point $C$ too, the pic looks ugly
 A: This problem is naturally projective, so you are handicapping yourself by forbidding methods like poles/polars/harmonic cross-ratios. The following uses Menelaus' theorem -- you might categorise it as a "higher theorem", but it was known $2000$ years ago so...

Let $X=IP\cap BC$. Then $XI\cdot XP=XD^2$, so $X$ lies on the radical axis of $(AIP)$ and the incircle, which is $EF$. So now apply Menelaus' theorem on $\triangle ABC$ and line $XFE$ (undirecting lengths):
$$1=\frac{BX}{XC}\cdot\frac{CF}{FA}\cdot\frac{AE}{EB}=\frac{BX}{XC}\cdot\frac{s-c}{s-a}\cdot\frac{s-a}{s-b}\implies\frac{BX}{XC}=\frac{s-b}{s-c}=\frac{BD}{DC}.$$
Now by the sine rule, we deduce that
\begin{align*}
1=\frac{BD}{DC}\div\frac{BX}{XC}&=\frac{\sin\angle BPD}{\sin\angle DPC}\div\frac{\sin\angle BPX}{\sin\angle XPC} \\
&=\frac{\sin\angle BPD}{\sin\angle DPC}\div\frac{\cos\angle BPD}{\cos\angle DPC} \\
&=\frac{\tan\angle BPD}{\tan\angle DPC}.
\end{align*}
Hence it follows that $\angle BPD=\angle DPC$.
A: I started to develop your idea from the OP, now we already have an accepted, good solution, but let me submit one more, since there is in the given constellation the following interesting property, not used or mentioned in all other solutions from AoPS:




Lemma: Define $T=MF\cap NE$. Then the lines $MF$, $NE$, $AI$, $BC$ are concurrent in $T$.


This "coincidence" immediately solves the problem in the OP (respecting the idea from the OP) by symmetry, since from the lemma $M,N$ are symmetric w.r.t. the axis $AJIT$, so $\overset\frown{AM}=\overset\frown{AN}$, and the angles of interest are pointing to these arcs.

Proof of the lemma:
First of all, the lines $EF$, $PI$, $BC$ are concurrent in a point $S$.
Many linked proofs use this step, here is an other (and the same) proof. Consider the inversion $*$ centered in $I$ which invariates $(DEF)$. The the circle $\odot(IPD)$ is transformed in the line perpendicular in $D^*=D$ on the diameter $ID$, which is $BDC$. The circle $\odot(IEF)$ is transformed in the line $EF=E^*F^*$, perpendicular on the diameter $IA$ and passing through the fixed points $E,F$. So $S=P^*$ is on both lines, i.e. $I,P,S=P^*=BC\cap EF$ are collinear.
Consider now the polygon $MFAENP$. Using Pascal's theorem, the following points are collinear.
$$
\begin{aligned}
MF \cap EN &= T\ ,\\
FA \cap NP &= B\ ,\\
AE \cap PM &= C\ .&&\text{ So }T\in BC\ .
\end{aligned}
$$
Further consider the hexagon $IP N EF A$, and apply again
Pascal's theorem, obtaining the colinearity of:
$$
\begin{aligned} 
IP\cap EF &= S\ ,\\
PN\cap FA &= B\ ,\\
NE\cap AI &= (\text{ a point on $SB$ and $NE$ }) =T\ .\text{ So }T\in AI\ .
\end{aligned}
$$
$\square$
