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

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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

Figure

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 $EG$.

Thus:

$2 \angle FEC = \angle AEC$

and:

$2 \angle CEG = \angle CEB$

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.

$\blacksquare$


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:

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


Hence

\(\ds l_1 l_2 + m_1 m_2\) \(=\) \(\ds \paren {\cos \alpha - \cos \beta} \paren {\cos \alpha + \cos \beta} + \paren {\sin \alpha - \sin \beta} \paren {\sin \alpha + \sin \beta}\)
\(\ds \) \(=\) \(\ds \paren {\cos^2 \alpha - \cos^2 \beta} + \paren {\sin^2 \alpha - \sin^2 \beta}\) 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.

$\blacksquare$


Sources

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