Perpendicular Lines and Finding Angle Mistakes | Geometry | LevelKite

When two lines intersect, they form four angles around the point where they meet. Sometimes one of those angles is a right angle, which is an angle that measures exactly $90^\circ$ . Those lines get a special name.

Definition. Two lines are perpendicular if they intersect and form a right angle. You write $\overleftrightarrow{AB} \perp \overleftrightarrow{CD}$ and say "line $AB$ is perpendicular to line $CD$ ."

A definition works in both directions. A right angle where the lines meet tells you the lines are perpendicular, and perpendicular lines tell you a right angle is there. In this lesson, $m\angle AEC$ means the measure of $\angle AEC$ .

The small square at

E

gives

\angle AEC

as a right angle. No other angle is marked, and nothing is read from the look of the drawing.

Worked example

From one right angle to four

Given: lines $AB$ and $CD$ intersect at $E$ , with $E$ between $A$ and $B$ and between $C$ and $D$ . The square mark in the figure above gives $\angle AEC$ as a right angle. To prove: the lines are perpendicular, and the other three angles at $E$ are right angles.

Apply the definition of a right angle. The square mark gives angle AEC as a right angle, so its measure is exactly 90 degrees.
$m\angle AEC = 90^\circ$
Apply the definition of perpendicular lines in the direction from right angle to perpendicular. The lines intersect and form a right angle, so they are perpendicular.
$\overleftrightarrow{AB} \perp \overleftrightarrow{CD}$
Angles AEC and BEC share ray EC, and rays EA and EB are opposite because E is between A and B, so they form a linear pair. By the Linear Pair Postulate, their measures add to 180 degrees. Substitute 90 degrees for the measure of angle AEC, then subtract 90 degrees from both sides.
$90^\circ + m\angle BEC = 180^\circ \;\Rightarrow\; m\angle BEC = 90^\circ$
Rays EB and EA are opposite, and rays ED and EC are opposite, so angles BED and AEC are vertical angles. By the Vertical Angles Theorem they are congruent, which means equal in measure.
$m\angle BED = m\angle AEC = 90^\circ$
In the same way, angles AED and BEC are vertical angles, so the Vertical Angles Theorem applies again.
$m\angle AED = m\angle BEC = 90^\circ$
Check the conditions. Each measure is 90 degrees, which is greater than 0 and at most 180, so no value is rejected. By the definition of a right angle, each of the four angles is a right angle.
$0 < 90 \le 180$

Answer

Lines $AB$ and $CD$ are perpendicular, and all four angles at $E$ measure $90^\circ$ , so all four are right angles.

Start from any right angle where two lines meet, and the same steps give the other three. So you have proved a theorem, a statement proved from definitions and postulates. Call it the Perpendicular Lines Theorem: if two lines are perpendicular, then all four angles they form are right angles. It follows that perpendicular lines form a linear pair of congruent angles.

The converse of an if-then statement swaps its "if" part and its "then" part. A true statement can have a false converse, so a converse needs its own proof. Swapping the last statement above gives this converse: if two intersecting lines form a linear pair of congruent angles, then the lines are perpendicular.

Here is the proof. Congruent angles have equal measures, so let each measure $y^\circ$ . The Linear Pair Postulate gives $y + y = 180$ , so $2y = 180$ . The Division Property of Equality gives $y = 90$ , which is greater than $0$ and at most $180$ . So each angle of the pair is a right angle, and the definition of perpendicular lines applies.

Key concept

Perpendicular lines, in both directions

Let lines $AB$ and $CD$ intersect at $E$ . They are perpendicular exactly when $\angle AEC$ is a right angle.

Perpendicular Lines Theorem: if the lines are perpendicular, then all four angles at $E$ are right angles.
Converse: if the lines form a linear pair of congruent angles, then they are perpendicular.

Use the theorem when you know the lines are perpendicular. Use the converse, or the definition, when you must show that they are. Each starting fact must be marked, stated in the text or already proved, because a figure is not drawn to scale. Each solved measure must be greater than $0^\circ$ and at most $180^\circ$ .

\overleftrightarrow{AB} \perp \overleftrightarrow{CD} \iff m\angle AEC = 90^\circ

Mistake: "It looks square, so it is $90^\circ$ ." In the figure below, lines $AB$ and $CD$ cross at $E$ with no square mark. A student sees a square-looking corner, writes $m\angle BEC = 90^\circ$ and calls the lines perpendicular. This reason fails because a figure is not drawn to scale. The same drawing could come with the stated fact $m\angle BEC = 88^\circ$ , since an $88^\circ$ angle can look just like a $90^\circ$ angle. Then the Linear Pair Postulate and the Vertical Angles Theorem give four angles of $88^\circ$ , $92^\circ$ , $88^\circ$ and $92^\circ$ , all greater than $0^\circ$ and at most $180^\circ$ . None is a right angle, so those lines would not be perpendicular. A missing square proves nothing either: only marks, stated facts and reasons decide.

Angle

BEF

measures

(2x+10)^\circ

and angle

FEC

measures

(3x-5)^\circ

. No square is drawn and the figure is not to scale, so any right angle at

E

must be proved from the stated facts.

Worked example

Identify perpendicular lines, then use them

Given: lines $AB$ and $CD$ intersect at $E$ , with $E$ between $A$ and $B$ and between $C$ and $D$ . Point $F$ lies in the interior of $\angle BEC$ , which means inside the angle, between its two rays. It is stated that $\angle AEC \cong \angle BEC$ , $m\angle BEF = (2x+10)^\circ$ and $m\angle FEC = (3x-5)^\circ$ . Decide whether the lines are perpendicular, then find both measures.

Identify the relationship. Angles AEC and BEC share ray EC, and rays EA and EB are opposite, so they form a linear pair. The problem states that they are congruent. By the converse, from a congruent linear pair to perpendicular lines, the lines are perpendicular.
$\overleftrightarrow{AB} \perp \overleftrightarrow{CD}$
Use the Perpendicular Lines Theorem in its forward direction, from perpendicular lines to right angles. Angle BEC is one of the four angles at E, so it is a right angle, and the definition of a right angle gives its measure.
$m\angle BEC = 90^\circ$
Point F lies in the interior of angle BEC, so the Angle Addition Postulate applies: the measures of the two smaller angles add to the measure of angle BEC.
$m\angle BEF + m\angle FEC = m\angle BEC$
Use the Substitution Property of Equality to replace each measure with its stated expression and angle BEC with 90. Then combine like terms. A total of 90 means these two angles are complementary.
$(2x+10)+(3x-5)=90 \;\Rightarrow\; 5x+5=90$
By the Subtraction and Division Properties of Equality, subtract 5 from both sides, then divide both sides by 5.
$5x=85 \;\Rightarrow\; x=17$
Substitute 17 into each expression and check. Both measures are greater than 0 and at most 180, and they add to 90.
$m\angle BEF = 2(17)+10 = 44^\circ,\quad m\angle FEC = 3(17)-5 = 46^\circ$

Answer

Lines $AB$ and $CD$ are perpendicular, with $m\angle BEF = 44^\circ$ and $m\angle FEC = 46^\circ$ .

Now use these facts to check someone else's work. A solution can look tidy and still be wrong, so test each line and not only the final answer. You can run the checks below in any order. When one fails, find the first line where the work goes wrong. Later lines can be correct algebra and still give a wrong answer, because they depend on that line.

These checks rely on the same conditions as before: a figure is not drawn to scale, and every angle measure must be greater than $0^\circ$ and at most $180^\circ$ .

Checks for work on angle relationships

Check	Question to ask	Mistake it catches
Givens	Is the fact marked or stated in the text, not just suggested by the drawing?	Calling an angle $90^\circ$ because it looks square
Relationship	Are the angles vertical, a linear pair, or parts of a larger angle?	Using a total of $90^\circ$ for a linear pair
Reason	Does the named reason prove this conclusion, in this direction?	Calling lines perpendicular because vertical angles are congruent
Algebra	Is each step applied to both sides of the equation?	Changing only one side of an equation
Values	Is each measure greater than $0^\circ$ and at most $180^\circ$ ?	Keeping a measure such as $-10^\circ$

Worked example

Find the mistake: the wrong total

Given: lines $AB$ and $CD$ intersect at $E$ , with $E$ between $A$ and $B$ and between $C$ and $D$ . No right angle is marked or stated. $m\angle AEC = (x-50)^\circ$ and $m\angle BEC = (2x+20)^\circ$ . A student writes: "The lines are perpendicular, so $(x-50)+(2x+20)=90$ . Then $3x-30=90$ , $3x=120$ and $x=40$ , so $m\angle AEC=-10^\circ$ and $m\angle BEC=100^\circ$ ." Find the mistake and correct the work.

Start with the values check. Angle measures must be greater than 0 and at most 180, and the student's value for angle AEC is negative. Reject that value, then look for the line that caused it.
$m\angle AEC = 40 - 50 = -10^\circ$
Check the givens. No mark or statement gives a right angle here, so the claim that the lines are perpendicular is only an assumption. It cannot serve as a reason.
Check the relationship. Angles AEC and BEC form a linear pair, since E is between A and B. So the Linear Pair Postulate gives a total of 180 degrees, whether or not the lines are perpendicular. The total of 90 degrees in the student's first line is the mistake. The algebra after it is correct but starts from a wrong equation.
$m\angle AEC + m\angle BEC = 180^\circ$
Use the Substitution Property of Equality to replace each measure with its stated expression, then combine like terms.
$(x-50)+(2x+20)=180 \;\Rightarrow\; 3x-30=180$
By the Addition and Division Properties of Equality, add 30 to both sides, then divide both sides by 3.
$3x=210 \;\Rightarrow\; x=70$
Substitute 70 and check. Both measures are greater than 0 and at most 180, and they add to 180. Use the Perpendicular Lines Theorem in its forward direction: if the lines were perpendicular, every angle at E would be a right angle. Neither measure is 90, so the lines are not perpendicular.
$m\angle AEC = 70-50 = 20^\circ,\quad m\angle BEC = 2(70)+20 = 160^\circ$

Answer

The mistake is the student's first line: it assumes perpendicular lines and uses $90^\circ$ as the total for a linear pair. In fact $x = 70$ , $m\angle AEC = 20^\circ$ , $m\angle BEC = 160^\circ$ , and the lines are not perpendicular.

Worked example

Find the mistake: the wrong reason

Given: lines $AB$ and $CD$ intersect at $E$ , with $E$ between $A$ and $B$ and between $C$ and $D$ . $m\angle AEC = (3x+15)^\circ$ and $m\angle BED = (5x-15)^\circ$ . A student writes: "Vertical angles are congruent, so $3x+15=5x-15$ . Then $x=15$ , and both angles measure $60^\circ$ . Two angles are congruent, so the lines are perpendicular." Find the mistake, then decide whether the lines are perpendicular.

Check the algebra. By the Subtraction, Addition and Division Properties of Equality, subtract 3x from both sides, add 15 to both sides, then divide both sides by 2. The student's value of x is correct.
$3x+15=5x-15 \;\Rightarrow\; 15=2x-15 \;\Rightarrow\; 30=2x \;\Rightarrow\; x=15$
Check the values. Substituting 15 gives 60 degrees for each angle, which is greater than 0 and at most 180, so the numbers pass.
$3(15)+15=60^\circ,\quad 5(15)-15=60^\circ$
Check the reason in the last sentence. Angles AEC and BED are vertical angles, and the Vertical Angles Theorem makes vertical angles congruent for any two intersecting lines. So this congruence cannot show that the lines are perpendicular. The converse needs a congruent linear pair, and these two angles are not a linear pair. This is the mistake.
Now test a linear pair. Angles AEC and BEC form a linear pair, so by the Linear Pair Postulate their measures add to 180 degrees. Subtract 60 degrees from both sides. The result, 120 degrees, is greater than 0 and at most 180.
$60^\circ + m\angle BEC = 180^\circ \;\Rightarrow\; m\angle BEC = 120^\circ$
Decide. Use the Perpendicular Lines Theorem in its forward direction: if the lines were perpendicular, every angle at E would be a right angle. Angle AEC measures 60 degrees, not 90, so the lines are not perpendicular.
$m\angle AEC = 60^\circ \ne 90^\circ$

Answer

The algebra is correct, but the last sentence is wrong: congruent vertical angles do not show perpendicular lines. Here $m\angle AEC = 60^\circ$ and $m\angle BEC = 120^\circ$ , so the lines are not perpendicular.

Takeaway: Two lines are perpendicular exactly when they intersect and form a right angle, and then all four angles they form are right angles. Base that claim on marks, stated facts and named reasons, never on how a figure looks, because a figure is not drawn to scale. When you check angle work, test each given, relationship, reason, algebra step and value, and reject any value that makes an angle measure zero, negative or greater than $180^\circ$ .