5.5: Perpendicular is shortest
Assume \(Q\) is the foot point of \(P\) on the line \(\ell\). Then the inequality
\(PX > PQ\)
holds for any point \(X\) on \(\ell\) distinct from \(Q\).
If \(P, Q\), and \(\ell\) are as above, then \(PQ\) is called the distance from \(P\) to \(\ell\).
- Proof
-
If \(P \in \ell\), then the result follows since \(PQ = 0\). Further we assume that \(P \not\in \ell\).
Let \(P'\) be the reflection of \(P\) across the line \(\ell\). Note that \(Q\) is the midpoint of \([PP']\) and \(\ell\) is the perpendicular bisector of \([PP']\). Therefore
\(PX = P'X\) and \(PQ = P'Q = \dfrac{1}{2} \cdot PP'\)
Note that \(\ell\) meets \([PP']\) only at the point \(Q\). Therefore, \(X \not\in [PP']\); by triangle inequality and Corollary 4.4.1 ,
\(PX + P'X > PP'\)
and hence the result: \(PX > PQ\).
Assume \(\angle ABC\) is right or obtuse. Show that
\(AC > AB.\)
- Hint
-
If \(\angle ABC\) is right, the statement follows from Lemma \(\PageIndex{1}\). Therefore, we can assume that \(\angle ABC\) is obtuse.
Draw a line \((BD)\) perpendicular to \((BA)\). Since \(\angle ABC\) is obtuse, the angles \(DBA\) and \(DBC\) have opposite signs.
By Corollary 3.4.1 , \(A\) and \(C\) lies on opposite sides of \((BD)\). In particular, \([AC]\) intersects \((BD)\) at a point; denote it by \(X\).
Note that \(AX < AC\) and by Lemma \(\PageIndex{1}\), \(AB \le AX\).
Suppose that \(\triangle ABC\) has right angle at \(C\). Show that for any \(X \in [AC]\) the distance from \(X\) to \((AB)\) is smaller than \(AB\).
- Hint
-
Let \(Y\) be the foot point of \(X\) on \((AB)\). Apply Lemma \(\PageIndex{1}\) to show that \(XY < AX \le AC < AB\).