# Constructing an Angle of 90°

Suppose that you have a line L and some point A on L:

How can you construct the perpendicular line to L through A? The steps of the construction are outlined below:

**Step 1:** Taking A as center and any radius, draw two circular arcs which intersect L on both sides of A (at B and C):

**Step 2:**** **Taking B and C as centers and a radius equal to more than half of BC, draw two arcs on the same side of L, which intersect each other at D.

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

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

1. AB = AC (arcs of equal radii)

2. BD = CD (arcs of equal radii)

3. AD = AD (common)

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

\(\angle BAD\) = \(\angle CAD\) = ½ (180^{0}) = 90^{0}

Thus, AD is perpendicular to L.

Note that you can construct an angle of 45^{0} by bisecting an angle of 90^{0}, and you can further construct an angle of 22.5^{0} by bisecting an angle of 45^{0}.