# How do you construct perpendicular bisectors of a triangle?

Apr 2, 2016

Draw two circles of the the same radius equal to the length of given segment $A B$ with centers at $A$ and $B$. They intersect at two points $P$ and $Q$. Line $P Q$ is a perpendicular bisector to $A B$.

#### Explanation:

To construct perpendicular bisectors of a triangle $\Delta A B C$ you have to consider each side separately as a segment ($A B$, $B C$ and $A C$) and construct a perpendicular bisector to each of them.

The easy way to construct a perpendicular bisector $P Q$ to segment $A B$ is pictured below.

Here the centers of these circles are the endpoints of a given segment $A B$ and their radiuses must be the same. The only condition is for these circles is the existence two points of intersection, $P$ and $Q$. For this the radius can be any, as long as it's greater than half of the length of $A B$. The simple method is to choose it to be equal to the length of $A B$.

What's more interesting is to prove that this construction delivers the perpendicular bisector.
Here is the proof.

Assume that $M$ is an intersection of $A B$ and $P Q$.
$A P = B P = A Q = B Q$ - each is a radius, which we have chosen
$\Delta A P Q = \Delta B P Q$ - by side-side-side theorem

Hence:
$\implies \angle A P Q = \angle B P Q$ as angles of congruent triangles lying across congruent sides $A Q$ and $B Q$
$\implies \Delta A P M = \Delta B P M$ by side-angle-side theorem
$\implies A M = B M$ as sides of congruent triangles lying across congruent angles $\angle A P M = \angle B P M$
$\implies \angle A M P = \angle B M P$ as angles of congruent triangles lying across congruent sides $A P$ and $B P$
$\implies \angle A M P = \angle B M P = {90}^{o}$ since their sum is ${180}^{o}$

So, we have proven that $M$ is a modpoint of $A B$ and $P M \bot A B$.