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
Best Answer
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 $\angle BPD=\angle DPC$.