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 \(\Delta CAO\):
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 (1800) = 900