Angles and parallel lines, with proofs

At this level geom­e­try changes char­ac­ter. It is no longer enough to mea­sure an angle with a pro­trac­tor and say "it looks like $60°$": we want to prove that it must be $60°$, using facts we already accept. This les­son revises the angle facts about inter­sect­ing lines, par­al­lel lines and tri­an­gles, and shows how each one is proved from sim­pler facts. Those proofs are the pat­tern for every­thing else in this course.

Axioms, the­o­rems and proofs#

An axiom is a basic fact we accept with­out proof because it is obvi­ously true, such as "a straight angle mea­sures $180°$". A the­o­rem is a state­ment we prove using axioms and ear­lier the­o­rems. A proof is the chain of rea­sons that leads from what is given to what we want, with a rea­son for every step.

Lin­ear pairs#

When a ray stands on a straight line, it makes two adja­cent angles on that line. Such a pair is called a lin­ear pair.

Lin­ear pair axiom. If a ray stands on a line, the two adja­cent angles it forms add up to $180°$. Con­versely, if two adja­cent angles add up to $180°$, their outer arms form a straight line.

Exam­ple 1#

Two angles form­ing a lin­ear pair are in the ratio $4 : 5$. Find them.

Let the angles be $4k$ and $5k$. Then $4k + 5k = 180°$, so $9k = 180°$ and $k = 20°$. The angles are $80°$ and $100°$.

Ver­ti­cally oppo­site angles#

When two lines cross, they make four angles. The angles oppo­site each other at the cross­ing point are called ver­ti­cally oppo­site angles.

The­o­rem. If two lines inter­sect, the ver­ti­cally oppo­site angles are equal.

Proof. Let lines $AB$ and $CD$ cross at $O$. Ray $OC$ stands on line $AB$, so by the lin­ear pair axiom

$$\angle AOC + \angle BOC = 180°$$

Ray $OB$ stands on line $CD$, so

$$\angle BOC + \angle BOD = 180°$$

Both sums equal $180°$, so $\angle AOC + \angle BOC = \angle BOC + \angle BOD$. Tak­ing $\angle BOC$ from both sides gives $\angle AOC = \angle BOD$. In the same way $\angle BOC = \angle AOD$.

Exam­ple 2#

Lines $AB$ and $CD$ inter­sect at $O$, with $\angle AOC = (3x + 10)°$ and $\angle BOC = (2x - 20)°$. Find all four angles.

$\angle AOC$ and $\angle BOC$ form a lin­ear pair:

$$\begin{aligned} (3x + 10) + (2x - 20) &= 180 \\ 5x - 10 &= 180 \\ x &= 38 \end{aligned}$$

So $\angle AOC = 3(38) + 10 = 124°$ and $\angle BOC = 2(38) - 20 = 56°$. By ver­ti­cally oppo­site angles, $\angle BOD = 124°$ and $\angle AOD = 56°$.

Par­al­lel lines and a trans­ver­sal#

A line that cuts two other lines at two dif­fer­ent points is called a trans­ver­sal. It makes eight angles, which come in named pairs.

Let a trans­ver­sal $EF$ cut line $AB$ at $P$ and line $CD$ at $Q$, with $E$ above $AB$, $F$ below $CD$, and $A$, $C$ on the left.

• Cor­re­spond­ing angles sit in match­ing posi­tions at the two cross­ings, such as $\angle EPB$ and $\angle PQD$.
• Alter­nate inte­rior angles lie between the two lines on oppo­site sides of the trans­ver­sal, such as $\angle APQ$ and $\angle PQD$.
• Inte­rior angles on the same side (co-inte­rior angles) lie between the lines on the same side of the trans­ver­sal, such as $\angle BPQ$ and $\angle PQD$.

When $AB \parallel CD$: cor­re­spond­ing angles are equal, alter­nate inte­rior angles are equal, and inte­rior angles on the same side add up to $180°$. Each state­ment also works back­wards: if any one of them holds, the lines are par­al­lel.

We accept the first (cor­re­spond­ing angles) as an axiom. The oth­ers fol­low from it with the lin­ear pair axiom and ver­ti­cally oppo­site angles. Another con­se­quence: two lines that are each par­al­lel to a third line are par­al­lel to each other.

Exam­ple 3#

With the trans­ver­sal described above, $AB \parallel CD$ and $\angle EPB = 65°$. Find $\angle PQD$, $\angle APQ$ and $\angle BPQ$.

$\angle PQD = \angle EPB = 65°$ (cor­re­spond­ing angles).

$\angle APQ = \angle PQD = 65°$ (alter­nate inte­rior angles).

$\angle BPQ = 180° - \angle EPB = 115°$ (lin­ear pair on line $EF$).

Check: $\angle BPQ + \angle PQD = 115° + 65° = 180°$, as co-inte­rior angles should.

Exam­ple 4#

$AB \parallel CD$, with $A$ and $C$ on the left and $B$ and $D$ on the right. A point $O$ lies between the two lines, to the left of $B$ and $D$. If $\angle ABO = 40°$ and $\angle CDO = 35°$, find $\angle BOD$.

The trick is to draw a help­ing line. Through $O$ draw $OY$ par­al­lel to $AB$, point­ing to the right. Then $OY$ is par­al­lel to $CD$ as well.

Trans­ver­sal $OB$ cuts the par­al­lels $AB$ and $OY$, so $\angle BOY = \angle ABO = 40°$ (alter­nate angles). Trans­ver­sal $OD$ cuts $OY$ and $CD$, so $\angle DOY = \angle CDO = 35°$ (alter­nate angles).

$$\angle BOD = \angle BOY + \angle DOY = 40° + 35° = 75°$$

The angles of a tri­an­gle#

Angle sum prop­erty. The three angles of a tri­an­gle add up to $180°$.

Proof. Take tri­an­gle $ABC$. Through $A$ draw the line $PQ$ par­al­lel to $BC$, with $P$ on the side of $B$. Then $\angle PAB = \angle ABC$ and $\angle QAC = \angle ACB$ (alter­nate angles). The three angles $\angle PAB$, $\angle BAC$, $\angle QAC$ make the straight angle at $A$, so

$$\angle ABC + \angle BAC + \angle ACB = \angle PAB + \angle BAC + \angle QAC = 180°$$

Exte­rior angle the­o­rem. If a side of a tri­an­gle is extended, the exte­rior angle formed equals the sum of the two inte­rior oppo­site angles.

Proof. Extend $BC$ of tri­an­gle $ABC$ to $D$. Then $\angle ACD + \angle ACB = 180°$ (lin­ear pair) and $\angle A + \angle B + \angle ACB = 180°$ (angle sum). Com­par­ing, $\angle ACD = \angle A + \angle B$.

Exam­ple 5#

The angles of a tri­an­gle are in the ratio $2 : 3 : 4$. Find them.

$2k + 3k + 4k = 180°$, so $k = 20°$. The angles are $40°$, $60°$ and $80°$.

Exam­ple 6#

An exte­rior angle of a tri­an­gle is $110°$ and one of the inte­rior oppo­site angles is $45°$. Find the other two angles of the tri­an­gle.

The other inte­rior oppo­site angle is $110° - 45° = 65°$. The inte­rior angle next to the exte­rior angle is $180° - 110° = 70°$. Check: $45° + 65° + 70° = 180°$.

Prac­tice#

1. Two angles form­ing a lin­ear pair dif­fer by $40°$. Find them.
2. Two lines inter­sect and one of the angles formed is $47°$. Find the other three.
3. The angles of a lin­ear pair are $(2x + 15)°$ and $(3x - 5)°$. Find $x$ and both angles.
4. Two par­al­lel lines are cut by a trans­ver­sal. Two inte­rior angles on the same side are $(3y + 20)°$ and $(2y + 10)°$. Find both angles.
5. Two par­al­lel lines are cut by a trans­ver­sal, and a pair of cor­re­spond­ing angles are $(4x - 10)°$ and $(2x + 30)°$. Find the angle.
6. The angles of a tri­an­gle are in the ratio $3 : 5 : 7$. Find them.
7. An exte­rior angle of a tri­an­gle is $120°$, and the two inte­rior oppo­site angles are in the ratio $1 : 2$. Find all three angles of the tri­an­gle.
8. In tri­an­gle $ABC$, $\angle A = \angle B + \angle C$. Find $\angle A$.
9. In the arrange­ment of Exam­ple 4, $\angle ABO = 50°$ and $\angle CDO = 30°$. Find $\angle BOD$.

Answers#