Playfair's perpendicular counterpart.

Question

I am trying to prove that given any line $l$ and a point $A$ not on it, there exists a unique line $k$ that crosses $A$ and is perpendicular to $l$.

Proof: By the Expansion Postulate, I know there exists at least two distinct points on $l$, call them $B$ and $C$. By the line postulate, construct line $\overleftrightarrow{AB}$. By the protractor postulate, construct ray starting at $A$ and ending at point $D$ so that $\overrightarrow{AD}$ is coplanar to plane $ABC$ (plane postulate), so that $\measuredangle BAD = 180-2\measuredangle ABC$, and finally, so that $\overrightarrow{AD}\cap\overleftrightarrow{BC}\neq \emptyset$. Let $$\overrightarrow{AD}\cap\overleftrightarrow{BC}=E$$ by the line intersection postulate.

It is clear that $\triangle ABE$ is isosceles (isosceles triangle theorem). Now, by the protractor postulate, choose point $F$ in the interior of $\angle BAE$ so that $$\measuredangle BAF = \measuredangle FAE =\frac{1}{2} \measuredangle BAE$$ ... Therefore, $\overleftrightarrow{AF}$ is a perpendicular to $\overleftrightarrow{CB}=l$. $\blacksquare$

The rest of the proof, including uniqueness is not too bad. But overall, this felt incredibly long winded and inefficient. The high-school book I am using says that they are using the postulates from Euclid but after research it seems as though they are much more closely using Hilbert's Axioms.

My question is this, is this route of proof valid? What is a better proof? I am likely missing something quite simple. I am aware of this and this post but these do not assume Playfair's postulate which I have no issue accepting (even though I guess I haven't used it here.) Any guidance or critiques are much appreciated.

EDIT

Second proof: By the expansion postulate, let $l=\overleftrightarrow{BC}$. By protractor postulate, choose point $D$ on the same side of $l$ as $A$, so that $\measuredangle DBC=90^\circ$. By line postulate, construct $\overleftrightarrow{DB}$. Notice $l \perp \overleftrightarrow{DB}$. If $A\in \overleftrightarrow{DB}$ then $\overleftrightarrow{DB}$ is the perpendicular we are looking for. Otherwise, by playfair we have a unique line $k$ going through $A$ so that $k \parallel \overleftrightarrow{DB}$. Notice $k\cap l \neq \emptyset$ for if that were true we would have that $l \parallel k \parallel \overleftrightarrow{DB}$ but $l$ and $\overleftrightarrow{DB}$ share point $B$. So let $k \cap l = E$. From here, using the consecutive interior angle theorem, it is easy to see that $\angle BEA$ is right so that line $k$ is the perpendicular we need. $\blacksquare$

Answer 1

I'm going to write a proof for the existence, and then have a look on your proof.

Proof :

There is a point on $l$. Let us call it $B$.

Then, there is a point $C$ on the line $\overleftrightarrow{AB}$ such that $B$ is between $A$ and $C$.

Next, let us draw a circle $C_1$ with center $A$ and radius equal to $|AC|$. Then, $B$ is inside $C_1$ since we have $|AB|\lt |AC|$.

Let us take two distinct points $D,E$ on $l$ satisfying $|BD|=|BE|=2|AC|$.

Since $|AD|\gt |BD|-|AB|\gt 2|AC|-|AC|=|AC|$, we see that $D$ is outside $C_1$. So, between $B$ and $D$, there is an intersection point of $l$ with $C_1$. Let us call the point $F$.

Similarly, between $B$ and $E$ which is outside $C_1$, there is an intersection point of $l$ with $C_1$. Let us call the point $G$.

We see that $\triangle{AFG}$ is isosceles. The rest is the same as yours.$\ \blacksquare$

some comments :

• I've tried to be rigorous (I hope above is rigorous enough), but it was difficult to guess what "tools" we are allowed to use.

• I was not sure if we could use, without a proof, the fact that any line passing through a point inside a circle has exactly two intersection points with the circle.

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You said in your proof that $\overrightarrow{AD}\cap\overleftrightarrow{BC}\neq \emptyset$, but If I'm not mistaken, I don't think it always holds.

For example, if $A(1,-\sqrt 3),B(0,0)$ and $C(2,0)$, then $AD$ is parallel to $BC$.

Maybe you meant $\measuredangle DAB = 180-2\measuredangle ABC$ instead of $\measuredangle BAD = 180-2\measuredangle ABC$, but I'm not sure.

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Added :

I believe I came up with a much better proof and I added it to the original post. It really relies on the protractor postulate which to my understanding says that given $\overrightarrow{XY}$ I can find point $Z$ so that $\measuredangle XYZ$ can be whatever real number I want. Is this correct to your understanding?

The proof you added looks very nice to me, and I think your understanding is correct.

I've spent a lot of time looking for another proof, but I couldn't find any short proof. Then, I found Geometry proofs for theorems and problems concerning parallel and perpendicular lines.. The OP of the question says

"draw a parallel line through the point not on the original line, then draw a line perpendicular to the newly drawn parallel line by using the protractor postulate, which then must also be perpendicular to the original line."

I found this approach interesting. I think that the key point of this proof is the very first step.

Because it feels too powerful for a postulate.

I agree with you. You may find Removing the protractor and SAS postulates interesting.