MHB Proving Equality of Angles in an Acute Triangle

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In the discussion, a user seeks to prove that in an acute triangle ABC, with points D and E on side BC such that BD=CE and D between B and E, the angles ABP and ACP are equal given certain conditions. The user expresses frustration over difficulties in creating a precise diagram and doubts about the validity of their approach involving coordinate calculations. They received assistance from another member, Opalg, which helped clarify their diagram and resolve their concerns. The user acknowledges the support received and feels relieved to have made progress. The conversation highlights the challenges of geometric proofs and the importance of collaboration in problem-solving.
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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 $\angle ABP=\angle ACP$.
 
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[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.
 
Last edited:
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?
 
anemone said:
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?
[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.
 
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)
 
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