# Unique perpendicular theorem

From a point lying on a line it is possible to draw only one line perpendicular to this line.

## Step 1

We will prove that it is possible to draw a perpendicular from any point to a line.

Consider point A and line BC; at that point A does not lie on line BC.

Proof of the theorem. Step 1

## Step 2

Connect point A with point B whereby angle ABC is obtained.

Let us draw angle PBC, which is equal in measure to angle ABC.

Proof of the theorem. Step 2

## Step 3

As ∠ РВС = ∠АВС, then the first angle can be superimposed on the second so that side BP will line up with side AB and BC will be the common side.

To perform such superposition is the same as bending the figure over line BC.

The point A of side AB will superimpose onto a certain point on side BP. Let us designate this point by letter А1.

Proof of the theorem. Step 3

## Step 4

Let is unfold the figure for the sake of visibility.

Then АА1 will intersect BC at a certain point K.

Proof of the theorem. Step 4

## Step 5

Consider triangles АКВ and ВКА1:

АВ=ВА1 – by condition;

∠АВК = ∠КВА1;

АК – the common side.

Δ АВК= Δ КВА1by SAS criterion.

According to the property of congruent triangles:

∠ АКВ = ∠ ВКА1

Angles АКВ and ВКА1 are adjacent angles. The sum of adjacent angles is 180°:

As this angles are congruent, then:

It follows herein:

Thus AK is a perpendicular.

As the point and the line were taken arbitrarily, it means that from any point to any line it is possible to draw a perpendicular.

Proof of the theorem. Step 5

## Step 6

Now let is prove that AK is the only perpendicular from point K to line BC.

Let us assume that through point A it is possible to draw another perpendicular АК1 to line BC.

Then the two lines AK and АК1 being perpendicular to line BC intersect at one point A. But it is impossible as according to the properties of perpendicular lines they cannot intersect. Consequently, from point A it is possible to draw only one perpendicular.

Unique perpendicular theorem is proved.

Proof of the theorem. Step 6