Suppose that you are given a line L, and a point A not on L**:**

How can you construct the perpendicular to L through A?

**Step 1:** Taking A as center and a sufficiently large radius, draw an arc which intersects L at two points, B and C, as shown below:

Can you see why the radius of this arc must be greater than a minimum value? What is that minimum value?

**Step 2:** Taking B and C as centers and a radius equal to more than ½ BC, draw two arcs which intersect each other at D:

**Step 3:** Draw a line through A and D. This is the required perpendicular.

**Proof:** Compare \(\Delta ABD\) and \(\Delta ACD\):

1. AB = AC (radii of the same circular arc)

2. BD = CD (arcs of the same radii)

3. AD = AD (common)

By the SSS criterion, the two triangles are congruent, which means that \(\angle BAO\) = \(\angle CAO\). Now, compare \(\Delta BAO\) and \(\Delta CAO\):

1. AB = AC (radii of the same circular arc)

2. \(\angle BAO\) = \(\angle CAO\) (shown above)

3. AO = AO (common)

By the SAS criterion, the two triangles are congruent, which means that

\(\angle BOA\) = \(\angle COA\) = ½ (180^{0}) = 90^{0}