In a plane if two lines are perpendicular to same line, prove that two lines are parallel.

From ProofWiki

In the words of Euclid:

(The Elements: Book $\text{XI}$: Proposition $6$)

Proof

Let $AB$ and $CD$ be two straight lines at right angles to the plane of reference.

It is to be demonstrated that $AB$ is parallel to $CD$.

Let $AB$ and $CD$ meet the plane of reference at $B$ and $D$ respectively.

Let the straight line $BD$ be joined.

Let $DE$ be drawn in the plane of reference at right angles to $BD$ such that $DE = AB$.

Let $BE$, $AE$ and $AD$ be joined.

We have that $AB$ is at right angles to the plane of reference.

So by Book $\text{XI}$ Definition $3$: Line at Right Angles to Plane:

But each of $BD$ and $BE$ is in the plane of reference and meets $AB$.

Therefore $\angle ABD$ and $\angle ABE$ are both right angles.

For the same reason $\angle CDB$ and $\angle CDE$ are both right angles.

We have that $AB = DE$, and that $BD$ is common.

Thus the two sides $AB$ and $BD$ of $\triangle ABD$ equal the two sides $BD$ and $BE$ of $\triangle EDB$.

The triangles $\triangle ABD$ and $\triangle EDB$ both include right angles.

Therefore by Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Equality:

We have that:

and:

Thus the two sides $AB$ and $BE$ of $\triangle ABE$ equal the two sides $ED$ and $DA$ of $\triangle EDA$.

We also have that $AE$ is common.

So from Proposition $8$ of Book $\text{I} $: Triangle Side-Side-Side Equality:

But $\angle ABE$ is a right angle.

Therefore $ED$ is at right angles to $DA$.

But $ED$ is also at right angles to the straight lines $BD$ and $DC$.

Therefore $ED$ is set up at right angles to the three straight lines $BD$, $DA$ and $DC$ at their meeting points.

Therefore from Proposition $5$ of Book $\text{XI} $: Three Intersecting Lines Perpendicular to Another Line are in One Plane:

But from Proposition $2$ of Book $\text{XI} $: Two Intersecting Straight Lines are in One Plane:

Therefore $AB$ is in the same plane as $DB$ and $DA$.

Therefore the straight lines $AB$, $BD$ and $DC$ are in one plane.

Also, each of $\angle ABD$ and $\angle BDC$ is a right angle.

Therefore from Supplementary Interior Angles implies Parallel Lines:

$\blacksquare$

Historical Note

This proof is Proposition $6$ of Book $\text{XI}$ of Euclid's The Elements.

Sources

Text Solution

Solution : Solution<br>Let lines `l` and `m` are perpendicular to `n`, then<br>`/_1=/_2=90^@`<br>See in the figure<br>Since, lines `l` and `m` cut by a transversal line `n` and the corresponding angles are equal, which shows that, line `l` is parallel to line `m`.

In figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

∵ Lines AB and CD intersect at O∴ ∠AOC = ∠BOD| Vertically Opposite AnglesBut ∠BOD = 40°    ...(1) | Given∴ ∠AOC = 40°    ...(2)Now, ∠AOC + ∠BOE = 70°⇒ 40° + ∠BOE = 70° | Using (2)⇒ ∠BOE = 70° - 40°⇒ ∠BOE = 30°Again,Reflex ∠COE= ∠COD + ∠BOD + ∠BOE= ∠COD + 40° + 30°| Using (1) and (2)= 180° + 40° + 30°| ∵ Ray OA stands on line CD|∴ ∠AOC + ∠AOD = 180° (Linear Pair Axiom) ⇒ ∠COD = 180°

= 250°.

If two straight lines are perpendicular to the same line, prove that they are parallel to each other.

Open in App