Proving Equality of Angles in an Acute Triangle

[anemone]

Given that $ABC$ is an acute triangle with $AC>AB$ and $D$ and $E$ be points on side $BC$ such that $BD=CE$ and $D$ lies between $B$ and $E$. Suppose there exists a point $P$ inside the triangle $ABC$ such that $PD$ is parallel to $AE$ and $\angle BAP=\angle CAE$.

Prove that $\angle ABP=\angle ACP$.

 

[anemone]

Spoiler: Solution of other

[TIKZ]\draw[thin] (0,0) circle (5cm);
\coordinate[label=left:B] (B) at (-4,3);
\coordinate[label=right:C] (C) at (6,3);
\coordinate[label=above:A] (A) at (-0.33,7);
\coordinate[label=above: P] (P) at (-1,4.9);
\coordinate[label=left: G] (G) at (-5,-0);
\coordinate[label=right: F] (F) at (3.8,-3.25);
\coordinate[label=above: D] (D) at (-0.27,3);
\coordinate[label=below: E] (E) at (2.2,3);
\draw (B) -- (C)-- (A)-- (B);
\draw (B) -- (P)-- (C)-- (B);
\draw (B) -- (G)-- (F)-- (B);
\draw (P) -- (F);
\draw (A) -- (E);
\draw (A) -- (P);
\draw (G) -- (P);
\draw (C) -- (F);
\begin{scope}
\path[clip] (B) -- (A) -- (P) -- cycle;
\draw[thick,red,double] (A) circle (1.3295);
\end{scope}
\begin{scope}
\path[clip] (B) -- (G) -- (P) -- cycle;
\draw[thick,red,double] (G) circle (1.3295);
\end{scope}
\begin{scope}
\path[clip] (E) -- (A) -- (C) -- cycle;
\draw[thick,red,double] (A) circle (1.3295);
\end{scope}
\begin{scope}
\path[clip] (B) -- (F) -- (P) -- cycle;
\draw[thick,red,double] (F) circle (1.3295);
\end{scope}
\begin{scope}
\path[clip] (A) -- (B) -- (P) -- cycle;
\draw[thick,blue] (B) circle (0.4177);
\end{scope}
\begin{scope}
\path[clip] (B) -- (P) -- (G) -- cycle;
\draw[thick,blue] (P) circle (0.4177);
\end{scope}
\begin{scope}
\path[clip] (A) -- (C) -- (P) -- cycle;
\draw[thick,blue] (C) circle (0.4177);
\end{scope}
\begin{scope}
\path[clip] (B) -- (F) -- (G) -- cycle;
\draw[thick,blue] (F) circle (0.4177);
\end{scope}
[/TIKZ]

Let $ABFC$ and $ABGP$ be parallelograms. Since $\triangle AEC \cong \triangle FDB$, we have $\angle BFP=\angle BFD=\angle CAE=\angle BAP=\angle BGP$. This implies $BGFP$ are concyclic.

Since $\triangle APC \cong \triangle BGF$, we have $\angle ABP=\angle BPG=\angle BFG=\angle ACP$ (Q.E.D.).

Remark: I know the parallelogram $ABFC$ looks distorted, but please cut me some slack (LOL), I have been trying my best to draw the diagram in TiKZ and some of my initial attempts work some never (which puzzled me much). I will continue to investigate the whys and eventually I will replace this diagram with a perfect one. Thanks to Klaas van Aarsen for his patience and constant help to guide me through some of the issues I encountered.

 

[anemone]

I spent much time to solve for each coordinates in order to get a more concise diagram to showcase how the above argument works. But in the process of doing so, I can't help but started to have doubt about the validity of the argument and less sure if it is a sound method.

Can someone please weigh in?

 

[Opalg]

I spent much time to solve for each coordinates in order to get a more concise diagram to showcase how the above argument works. But in the process of doing so, I can't help but started to have doubt about the validity of the argument and less sure if it is a sound method.

Can someone please weigh in?

Spoiler: Edited diagram

[TIKZ]\draw[thin] (0,0) circle (5cm);
\coordinate[label=left:B] (B) at (-4,3);
\coordinate[label=right:C] (C) at (8.5,3);
\coordinate[label=above:A] (A) at (0,7.9);
\coordinate[label=above: P] (P) at (-1,4.9);
\coordinate[label=left: G] (G) at (-5,0);
\coordinate[label=right: F] (F) at (4.625,-1.9);
\coordinate[label=above: D] (D) at (0,3);
\coordinate[label=below: E] (E) at (3.7,3);
\draw (B) -- (C)-- (A)-- (B);
\draw (B) -- (P)-- (C)-- (B);
\draw (B) -- (G)-- (F)-- (B);
\draw (P) -- (F);
\draw (A) -- (E);
\draw (A) -- (P);
\draw (G) -- (P);
\draw (C) -- (F);
\begin{scope}
\path[clip] (B) -- (A) -- (P) -- cycle;
\draw[thick,red,double] (A) circle (1.3295);
\end{scope}
\begin{scope}
\path[clip] (B) -- (G) -- (P) -- cycle;
\draw[thick,red,double] (G) circle (1.3295);
\end{scope}
\begin{scope}
\path[clip] (E) -- (A) -- (C) -- cycle;
\draw[thick,red,double] (A) circle (1.3295);
\end{scope}
\begin{scope}
\path[clip] (B) -- (F) -- (P) -- cycle;
\draw[thick,red,double] (F) circle (1.3295);
\end{scope}
\begin{scope}
\path[clip] (A) -- (B) -- (P) -- cycle;
\draw[thick,blue] (B) circle (0.4177);
\end{scope}
\begin{scope}
\path[clip] (B) -- (P) -- (G) -- cycle;
\draw[thick,blue] (P) circle (0.4177);
\end{scope}
\begin{scope}
\path[clip] (A) -- (C) -- (P) -- cycle;
\draw[thick,blue] (C) circle (0.4177);
\end{scope}
\begin{scope}
\path[clip] (B) -- (F) -- (G) -- cycle;
\draw[thick,blue] (F) circle (0.4177);
\end{scope} [/TIKZ]
Does this help at all? I have kept all your TikZ coding apart from changing some of the coordinates. I had to cheat on the position of $E$ because if I had made $BD=EC$ then $E$ would have been almost exactly on the circle, which would be misleading. I concentrated on making the parallelograms the right shape. That has the effect of showing that $PCFG$ is also a parallelogram (because the sides $GP$ and $FC$ are equal and parallel). That in turn explains why the triangles $APC$ and $BGF$ are congruent, which I previously found hard to see.

I am definitely happy about the validity of the argument.

 

[anemone]

Thanks Opalg for your help to salvage my diagram...(Handshake) I deeply appreciate it. I have tried so many different pairs of coordinates but still I couldn't get it to work. I even tried to solve for all the coordinates by at first letting coordinates $B, P, A$ to take on some values, and worked out the rest, using a bunch of formulas (distance and gradient, along with Sine and Cosine Rules) and in the end, I got a really ugly equation in two variables to solve. I guessed and checked the answer and still, I couldn't get to draw a perfect diagram to illustrate the whole thing. I tried again and again, exhausting all my energy and patience, and that was when I started to think if it is even possible to draw such a diagram. And I had to ask the members of MHB to help me out...

You just saved my day, Opalg! Without your help, I would still continue trying for more today and more hair tearing would ensue. Now, I can put this to rest. (Happy)