Let us draw a line m. Let A be a point.

Let us draw two intersecting lines l and p through the point A, such that they are perpendicular to the line m at points P and Q.

⇒ ∠P = 90°, and ∠Q = 90° - - - - (i)

We have to prove that through a given point, we can draw only one perpendicular to a given line.

⇒ To prove that ∠A = 0°

Now, we can see in ΔPAQ,

∠P + ∠A + ∠Q = 180° (Since the sum of the angles of a triangle is equal to 180°)

⇒ ∠A + 90° + 90° = 180° [From equation (i)]

⇒ ∠A + 180° = 180°

⇒ ∠A = 0°

This means that ∠A doesn’t exist, i.e., the lines l and p coincide with each other.

Therefore, it is true that through a given point, we can draw only one perpendicular to a given line.