# Proving Equality of Angles in an Acute Triangle

• MHB
• anemone
In summary, the conversation discusses a problem involving an acute triangle $ABC$, with points $D$ and $E$ on side $BC$ such that $BD=CE$, and a point $P$ inside the triangle. The problem is to prove that $\angle ABP=\angle ACP$, given that $PD$ is parallel to $AE$ and $\angle BAP=\angle CAE$. The participants of the conversation had doubts about the validity of their argument and asked for help from others.
anemone
Gold Member
MHB
POTW Director
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 $\angle ABP=\angle ACP$.

[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.

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.

[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)

## 1. How do you prove that the angles in an acute triangle are equal?

To prove that the angles in an acute triangle are equal, we can use the fact that the sum of the angles in any triangle is 180 degrees. Since an acute triangle has three angles that are less than 90 degrees, their sum must be less than 180 degrees. Therefore, each angle in an acute triangle must be less than 60 degrees. If we can show that all three angles are less than 60 degrees, then we can conclude that they are equal.

## 2. What is the theorem used to prove equality of angles in an acute triangle?

The theorem used to prove equality of angles in an acute triangle is the acute angle theorem. This theorem states that in an acute triangle, the smallest angle is opposite the shortest side, and the largest angle is opposite the longest side. This means that if we can show that the three angles are equal, then the three sides must also be equal, making it an equilateral triangle.

## 3. Can you prove equality of angles in an acute triangle without using the acute angle theorem?

Yes, there are other methods to prove equality of angles in an acute triangle. One method is to use the law of sines, which states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. By using this law, we can show that the three angles in an acute triangle must be equal.

## 4. What are some real-life applications of proving equality of angles in an acute triangle?

Proving equality of angles in an acute triangle is important in many fields, such as engineering, architecture, and navigation. In engineering and architecture, it is necessary to ensure that angles are equal in order to create stable and symmetrical structures. In navigation, knowing the angles of a triangle can help determine the location of a ship or plane.

## 5. Are there any other types of triangles where you can prove equality of angles?

Yes, there are other types of triangles where you can prove equality of angles. In an equilateral triangle, all three angles are equal to 60 degrees. In an isosceles triangle, the two angles opposite the equal sides are equal. In a right triangle, one angle is always equal to 90 degrees. In a scalene triangle, none of the angles are equal, but you can still use the law of sines to prove certain relationships between the angles.

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