Perpendicular bisectors and altitudes

The perpendicular bisector of a segment is the line through the segment’s midpoint perpendicular to it.

Theorem: the perpendicular bisector is the locus of all the points at equal distances from the ends of the segment.

Given: $\displaystyle M$ is the midpoint of $\displaystyle AB$, and $\displaystyle PM\perp AB$.
Prove: $\displaystyle {PA}={PB}$.

The proof below shows that the triangles $\displaystyle PMA$ and $\displaystyle PMB$ are congruent (SAS).

The other direction is true too: a point at equal distances from $\displaystyle A$ and from $\displaystyle B$ is the vertex of an isosceles triangle with the base $\displaystyle AB$, and in such a triangle the median to the base is also an altitude – so the point is on the perpendicular bisector.

In the triangle $\displaystyle ABC$, $\displaystyle MP$ is the perpendicular bisector of the side $\displaystyle AB$, and $\displaystyle NQ$ the perpendicular bisector of the side $\displaystyle AC$.
The two perpendicular bisectors cross at the point $\displaystyle O$.
Prove: $\displaystyle {OB}={OC}$ – that is, $\displaystyle O$ is on the perpendicular bisector of the side $\displaystyle BC$ as well.

Theorem: the three perpendicular bisectors of the sides of a triangle meet at one point, $\displaystyle O$, at equal distances from the three vertices (Exercise 1).

Where $\displaystyle O$ is:

• in an acute triangle – inside the triangle;
• in a right-angled triangle – at the midpoint of the hypotenuse: the median to the hypotenuse is half the hypotenuse, so the midpoint of the hypotenuse is at equal distances from the three vertices;
• in an obtuse triangle – outside the triangle.

The equation of the perpendicular bisector of the segment $\displaystyle AB$ comes in two steps: the point – the segment’s midpoint; the slope – its product with the slope of $\displaystyle AB$ is $\displaystyle -1$.

Example: $\displaystyle A(1,2)$ and $\displaystyle B(5,4)$.

• the midpoint: $\displaystyle M(3,3)$;
• $\displaystyle {m_{AB}}={\frac{4-2}{5-1}}={\frac{1}{2}}$, so the slope of the perpendicular bisector is $\displaystyle -2$;
• $\displaystyle {y-3}={-2(x-3)}$, that is $\displaystyle {y}={-2x+9}$.

The vertices of the triangle $\displaystyle ABC$ are $\displaystyle A(0,0)$, $\displaystyle B(8,0)$ and $\displaystyle C(2,6)$.
The board shows the perpendicular bisector of the side $\displaystyle AB$: $\displaystyle {x}={4}$.

Write on the board the equation of the perpendicular bisector of the side $\displaystyle AC$.

Find the point $\displaystyle O$ where the two perpendicular bisectors meet.

Find the distances of $\displaystyle O$ from the three vertices.
What do you get?

Theorem: the three altitudes of a triangle, or the lines containing them, meet at one point, $\displaystyle H$.

In an acute triangle the three altitudes are inside the triangle, and so is $\displaystyle H$:

In an obtuse triangle two altitudes fall on extensions of the sides, and $\displaystyle H$ is outside the triangle – the extensions of the altitudes meet:

In a right-angled triangle each leg is also the altitude to the other leg, so $\displaystyle H$ is the vertex of the right angle.

In the triangle of Exercise 2: $\displaystyle A(0,0)$, $\displaystyle B(8,0)$ and $\displaystyle C(2,6)$.
The altitude from $\displaystyle C$ to the side $\displaystyle AB$ is on the line $\displaystyle {x}={2}$.

Write on the board the equations of the lines containing the altitude from $\displaystyle A$ and the altitude from $\displaystyle B$.
(A hint: $\displaystyle {m_{BC}}={-1}$ and $\displaystyle {m_{AC}}={3}$.)

Find the point $\displaystyle H$ where the two altitudes cross.

Show that $\displaystyle H$ is on the altitude from $\displaystyle C$ as well.
Is $\displaystyle H$ inside the triangle? What does that say about the triangle’s angles?