Sides, Angles, and Shortest Distances

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Learning goals

Compare sides and opposite angles within the same triangle
Justify side and angle ordering using an exterior angle and indirect proof
Identify the hypotenuse and explain why it is the longest side
Determine the shortest segment from a point to a line

Match each angle to its opposite side

The side opposite an angle is the side that does not touch the angle’s vertex. In triangle $ABC$ , side $BC$ is opposite angle $A$ . Side $AC$ is opposite angle $B$ , and side $AB$ is opposite angle $C$ .

Here $AB$ without a bar denotes the length of segment $AB$ . Angle sizes are their measures, such as $m\angle ABC$ . We compare lengths with lengths and angle measures with angle measures.

In the diagram, angle $B$ measures $80^\circ$ and angle $C$ measures $40^\circ$ . The side facing $B$ looks longer than the side facing $C$ . Rotating the paper would change neither the angles nor that comparison. The two proofs below show why the comparison must hold.

Opposite means across from the vertex. The labels are givens; the apparent side lengths are not evidence.

All triangles in this lesson have three noncollinear vertices. A flat arrangement of points is not a triangle here.

Check your understanding

In triangle $KLM$ , angle $L$ is larger than angle $M$ . Which sides are opposite those angles, in that order?

A KL and KM B KM and KL C LM and KL D KM and LM

An exterior angle is larger

An exterior angle is larger than either of its remote interior angles, the two interior angles away from its vertex. For example, remote angles of $43^\circ$ and $62^\circ$ give an exterior angle of

43^\circ+62^\circ=105^\circ.

Their sum is greater than each positive part. This is the inequality consequence of the exterior angle theorem.

B, C, and D lie in that order on a line. Shading identifies the remote interior angles at A and B; it does not mark them equal.

In general, the exterior angle theorem gives

m\angle ACD=m\angle CAB+m\angle ABC.

Both remote interior angles have positive measure, so

m\angle ACD>m\angle CAB.

Also,

m\angle ACD>m\angle ABC.

The adjacent interior angle at $C$ is not one of the remote angles. An exterior angle need not be larger than that adjacent angle.

Why a longer side faces a larger angle

A longer side in a triangle faces a larger angle. We can prove this by cutting the longer side to make an isosceles triangle.

A longer side has a larger opposite angle#

First follow the construction with numbers. Let $D$ lie between $A$ and $C$ , and make $AD$ equal to $AB$ . Imagine an instance where angle $ABD$ is $70^\circ$ and angle $DBC$ is $25^\circ$ .

The cutoff AD matches AB. BD divides angle ABC, and DA extends DC through D. The numbers illustrate one instance, not additional assumptions in the general proof.

The equal sides in triangle $ABD$ make its base angles equal. Thus angle $BDA$ also measures $70^\circ$ .

Angle $BDA$ is exterior to triangle $BDC$ . Its remote angles at $B$ and $C$ add to $70^\circ$ , so

m\angle BCD=70^\circ-25^\circ.

Therefore,

m\angle BCD=45^\circ.

The full angle at $B$ combines the two parts:

m\angle ABC=70^\circ+25^\circ.

Thus angle $ABC$ measures $95^\circ$ , greater than angle $ACB$ , which measures $45^\circ$ . The larger side $AC$ faces the larger angle.

Now let $ABC$ be any triangle with

AC>AB.

Choose $D$ on segment $AC$ so that

AD=AB.

This point lies strictly between $A$ and $C$ : the positive length $AB$ is less than $AC$ . Join $B$ to $D$ . The diagram below keeps only the general construction.

The general construction keeps only the chosen equality and incidence. D is not assumed to be a midpoint.

Triangle $ABD$ is isosceles, so

m\angle ABD=m\angle BDA.

Since ray $DA$ extends ray $DC$ through $D$ , angle $BDA$ is exterior to triangle $BDC$ . The exterior angle inequality gives

m\angle BDA>m\angle BCD.

Because $D$ lies between $A$ and $C$ , segment $BD$ lies inside angle $ABC$ , so the full angle is greater than its part:

m\angle ABC>m\angle ABD.

The comparisons connect in this order:

\begin{aligned} m\angle ABC&>m\angle ABD,\\ m\angle ABD&=m\angle BDA,\\ m\angle BDA&>m\angle BCD,\\ m\angle BCD&=m\angle ACB. \end{aligned}

The last equality holds because rays $CA$ and $CD$ coincide. Therefore,

m\angle ABC>m\angle ACB.

Neither the isosceles step nor the exterior angle inequality needed the special numbers $70$ and $25$ . The argument works for every positive difference between the two side lengths.

Reverse the argument

A larger angle also faces a longer side.

A larger angle has a longer opposite side#

For a numerical test, suppose angle $B$ is $75^\circ$ and angle $C$ is $45^\circ$ . Could $AC$ and $AB$ be equal? That would force the two angles to be equal. Could $AC$ be shorter? The result just proved would force angle $B$ to be smaller. Both alternatives contradict the given measures.

Suppose

m\angle ABC>m\angle ACB.

We use indirect proof: rule out the alternatives to the desired comparison.

If $AC$ equaled $AB$ , the isosceles base angles at $B$ and $C$ would be equal. That contradicts the given strict inequality.

If $AC$ were less than $AB$ , the longer-side result would make angle $ACB$ greater than angle $ABC$ . That also contradicts the given.

Equality and the reverse inequality have both failed. The remaining possibility is

AC>AB.

The numerical test used $75$ and $45$ , but the two contradictions depend only on which angle is larger.

The two results apply within the same triangle: compare the sides by comparing their opposite angles, or compare the angles by comparing their opposite sides. Equal sides and equal opposite angles remain the equality case from the isosceles theorems.

Check your understanding

Triangle $ABC$ has

AC>AB.

In the cutoff proof, $D$ lies on $AC$ with

AD=AB.

Which fact connects the equal base angles of triangle $ABD$ to angle $C$ ?

A Angle BDA is exterior to triangle BDC B BD is perpendicular to AC C D is the midpoint of AC D Triangle BDC is equilateral

Check your understanding

Triangle $ABC$ has angle $B$ larger than angle $C$ . In an indirect proof, suppose

AC=AB.

Why does this contradict the given angle comparison?

A It would make angle A a right angle B It would make angles A and B equal C It would make angles B and C equal D It would make AC shorter than AB

Put comparisons in order

Worked example 1 Rank three sides

Problem. Triangle $PQR$ has angle $P$ of $48^\circ$ , angle $Q$ of $67^\circ$ , and angle $R$ of $65^\circ$ . List its sides from shortest to longest.

The given angle measures identify the opposite sides. Do not rank sides by their position on the page.

Solution. The smallest angle is at $P$ , so its opposite side $QR$ is shortest. The middle angle is at $R$ , so $PQ$ comes next. The largest angle is at $Q$ , so $PR$ is longest:

QR<PQ.

Then,

PQ<PR.

Worked example 2 Turn a geometric comparison into algebra

Problem. An existing triangle $ABC$ has angle $B$ larger than angle $C$ . Its side lengths are

AC=x+6,

and

AB=2x+1.

Given that $x$ is positive, find the necessary restriction on $x$ .

The labels show the given comparison and side expressions.

Solution. Side $AC$ faces angle $B$ , so

x+6>2x+1.

Subtract $x$ and then subtract $1$ :

5>x.

Combining this with the given positivity,

0<x<5.

At $x=5$ the sides would match, which would force equal opposite angles. We exclude that endpoint. This is a necessary restriction from the comparison; it does not specify the triangle’s third side.

The longest side in a right triangle

The side opposite the right angle is the hypotenuse. It is always the longest side of a right triangle.

For instance, a triangle with one right angle and an acute angle of $35^\circ$ has a remaining angle of

180^\circ-90^\circ-35^\circ=55^\circ.

The right angle is larger than both acute angles. Its opposite side is therefore longer than both other sides.

The hypotenuse is longer than either leg#

In any right triangle, the two other angles are positive and sum to $90^\circ$ . Each is therefore less than $90^\circ$ . The larger-angle result puts the side opposite the right angle ahead of each side opposite an acute angle.

Worked example 3 Find the hypotenuse from the right angle

Problem. Triangle $PQR$ is right at $Q$ . Angle $P$ measures $58^\circ$ . Identify the hypotenuse and order the sides from shortest to longest.

The square marks the given right angle at Q.

Solution. Side $PR$ is opposite $Q$ , so it is the hypotenuse and longest side. The angle sum gives

m\angle R=180^\circ-90^\circ-58^\circ.

Therefore,

m\angle R=32^\circ.

Angle $R$ is smallest, so its opposite side $PQ$ is shortest. Angle $P$ is next, so $QR$ is between the other two:

PQ<QR.

Also,

QR<PR.

Check your understanding

A right triangle $XYZ$ is right at $Y$ . Which side must be longest?

A XY B YZ C XZ D There is not enough information

The shortest route to a line

For a point outside a line, the perpendicular segment to the line is shorter than every other segment from that point to the line.

Suppose $PH$ meets line $\ell$ at a right angle. A segment to another point $Q$ on the line forms right triangle $PHQ$ . Its hypotenuse is $PQ$ , so $PQ$ is longer than $PH$ . A perpendicular segment of $8$ centimeters therefore beats every slanted segment to the line, even if none of their lengths is known.

Both slanted segments are hypotenuses of right triangles with the same perpendicular leg PH. Q can be on either side of H.

The perpendicular is the shortest segment to a line#

Let $P$ be outside line $\ell$ , and let $H$ be the foot of its perpendicular. Take any point $Q$ on $\ell$ other than $H$ .

Triangle $PHQ$ is a right triangle because $HQ$ lies on $\ell$ . Its hypotenuse is $PQ$ , so

PQ>PH.

This holds for every choice of $Q$ other than $H$ , on either side. The endpoint $H$ gives $PH$ itself, so $H$ is the unique closest point on the line.

The distance from a point to a line is the length of that perpendicular segment. If the point already lies on the line, its distance to the line is zero; there is no perpendicular segment of positive length from the point to the line, so no right triangle to use in that case.

Worked example 4 Check a claimed distance

Problem. Point $P$ is outside line $\ell$ , and $H$ is its perpendicular foot. Let $Q$ be another point on $\ell$ . The lengths, in centimeters, are

PH=8,

and

PQ=5x+3.

What restriction on $x$ is necessary?

The labels give the stated lengths in centimeters. The square marks the right angle at H.

Solution. Triangle $PHQ$ has hypotenuse $PQ$ . Its length must exceed the perpendicular length:

5x+3>8.

Subtract $3$ , then divide by positive $5$ :

x>1.

At the excluded endpoint $x=1$ , the slanted segment would have the same length as the perpendicular, contradicting the strict comparison.

Check your understanding

Point P is outside a line. PH is perpendicular to it and has length 7. Q is another point on the line. Which claimed length for PQ is impossible?

A 9 B 12 C 8 D 6

Sides, Angles, and Shortest Distances

Learning goals

Match each angle to its opposite side

An exterior angle is larger

Why a longer side faces a larger angle

Reverse the argument

Put comparisons in order

The longest side in a right triangle

The shortest route to a line

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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