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# 5.5: Perpendicular is shortest

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Lemma $$\PageIndex{1}$$

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: $$PX > PQ$$.

Exercise $$\PageIndex{1}$$

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 $$\PageIndex{1}$$, $$AB \le AX$$.

Exercise $$\PageIndex{2}$$

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 $$XY < AX \le AC < AB$$.