Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular

From ProofWiki
Jump to navigation Jump to search

Corollary to Two Angles on Straight Line make Two Right Angles

If a straight line meets another straight line, the bisectors of the two adjacent angles between them are perpendicular.

Proof 1

Let $AB$ and $CD$ be two straight lines that cross at $E$.

Let $\angle AEC$ be bisected by $EF$.

Let $\angle CEB$ be bisected by $EG$.


$2 \angle FEC = \angle AEC$


$2 \angle CEG = \angle CEB$

But from Two Angles on Straight Line make Two Right Angles, $\angle AEC + \angle CEB$ equal $2$ right angles.

Thus $2 \angle FEC + 2 \angle CEG$ equal $2$ right angles.

Hence $\angle FEG = \angle FEC + \angle CEG$ equals $1$ right angle.

That is $EF$ and $EG$ are perpendicular.


Proof 2

Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed in normal form as:

\(\ds \LL_1: \ \ \) \(\ds x \cos \alpha + y \sin \alpha\) \(=\) \(\ds p\)
\(\ds \LL_2: \ \ \) \(\ds x \cos \beta + y \sin \beta\) \(=\) \(\ds q\)

From Bisectors of Angles between Two Straight Lines, the angle bisectors of the angles formed at the point of intersection of $\LL_1$ and $\LL_2$ are given by:

\(\ds x \paren {\cos \alpha - \cos \beta} + y \paren {\sin \alpha - \sin \beta} - \paren {p - q}\) \(=\) \(\ds 0\)
\(\ds x \paren {\cos \alpha + \cos \beta} + y \paren {\sin \alpha + \sin \beta} - \paren {p + q}\) \(=\) \(\ds 0\)

These are in the form $l x + m y + n = 0$.

We use Condition for Straight Lines in Plane to be Perpendicular to prove that $l_1 l_2 + m_1 m_2 = 0$, where:

\(\ds l_1\) \(=\) \(\ds \cos \alpha - \cos \beta\)
\(\ds l_2\) \(=\) \(\ds \cos \alpha + \cos \beta\)
\(\ds m_1\) \(=\) \(\ds \sin \alpha - \sin \beta\)
\(\ds m_2\) \(=\) \(\ds \sin \alpha + \sin \beta\)


\(\ds l_1 l_2 + m_1 m_2\) \(=\) \(\ds \paren {\cos \alpha - \cos \beta} \paren {\cos \alpha + \cos \beta} + \paren {\sin \alpha - \sin \beta} \paren {\sin \alpha + \sin \beta}\)
\(\ds \) \(=\) \(\ds \paren {\cos^2 \alpha - \cos^2 \beta} + \paren {\sin^2 \alpha - \sin^2 \beta}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \paren {\cos^2 \alpha + \sin^2 \alpha} - \paren {\cos^2 \beta + \sin^2 \beta}\) rearranging
\(\ds \) \(=\) \(\ds 1 - 1\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds 0\)

Hence the result.