The circumscribed circle, the inscribed circle and the altitudes

Theorem: the three perpendicular bisectors of a triangle’s sides meet at one point. That point is at equal distances from the three vertices, so it is the centre of a circle through them – the circumscribed circle of the triangle.

Given: $\displaystyle O$ is on the perpendicular bisector of $\displaystyle AB$ (through its midpoint $\displaystyle M$) and on that of $\displaystyle BC$ (through its midpoint $\displaystyle N$); $\displaystyle K$ is the midpoint of $\displaystyle AC$.
To prove: $\displaystyle OK\perp AC$ – so $\displaystyle O$ is on the third perpendicular bisector too.

The proof below uses the perpendicular bisector as a locus, in both of its directions:

So every triangle can be inscribed in a circle, and through any three points not on one line passes a circle – exactly one: its centre must be the meeting point of the perpendicular bisectors.

The centre of the circumscribed circle depends on the triangle’s largest angle:

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

The legs of a right-angled triangle are $\displaystyle 6$ and $\displaystyle 8$.
Find the radius of its circumscribed circle.

The vertices of the triangle $\displaystyle ABC$ are $\displaystyle A(0,0)$, $\displaystyle B(8,0)$ and $\displaystyle C(2,6)$.

The perpendicular bisector of $\displaystyle AB$ is $\displaystyle {x}={4}$, and that of $\displaystyle AC$ is $\displaystyle {y}={-\frac{1}{3}x+\frac{10}{3}}$.
Find the centre $\displaystyle O$ of the circumscribed circle.

Write the equation of the circumscribed circle.

Theorem: the three angle bisectors of a triangle meet at one point. That point is at equal distances from the three sides, so it is the centre of a circle that touches the three sides – the inscribed circle of the triangle.

Why, by the angle bisector as a locus:

• $\displaystyle I$ is where the bisectors of the angles $\displaystyle A$ and $\displaystyle B$ meet;
• $\displaystyle I$ is on the bisector of $\displaystyle A$, so it is at equal distances from $\displaystyle AB$ and $\displaystyle AC$; it is on the bisector of $\displaystyle B$, so at equal distances from $\displaystyle AB$ and $\displaystyle BC$;
• so it is at equal distances from $\displaystyle AC$ and $\displaystyle BC$ – and it is on the bisector of $\displaystyle C$.

The radius of the inscribed circle is that common distance, $\displaystyle r$: the circle with centre $\displaystyle I$ and radius $\displaystyle r$ meets each side at the foot of the perpendicular from $\displaystyle I$, and every other point of the side is farther from $\displaystyle I$.

$\displaystyle I$ is the centre of the circle inscribed in the triangle $\displaystyle ABC$.
$\displaystyle {\angle B}={70^{\circ}}$ and $\displaystyle {\angle C}={50^{\circ}}$.
Find $\displaystyle \angle BIC$.

In the triangle $\displaystyle ABC$, $\displaystyle BI$ bisects the angle $\displaystyle B$ and $\displaystyle CI$ bisects the angle $\displaystyle C$.
Prove: $\displaystyle AI$ bisects the angle $\displaystyle A$ – write the proof above with claims and reasons.
(A hint: draw the perpendiculars from $\displaystyle I$ to the three sides.)

Theorem: the three altitudes of a triangle – or the lines containing them – meet at one point.

• In an acute-angled triangle the point is inside it; in a right-angled triangle it is the vertex of the right angle; in an obtuse-angled triangle it is outside it, where the lines of the altitudes meet.
• The proof is beyond this course; Exercise 5 shows it for one triangle, in coordinates.

The loci give the constructions:

• the perpendicular bisector of $\displaystyle AB$: two circles of the same radius, centred at $\displaystyle A$ and at $\displaystyle B$, meet at two points – each at equal distances from $\displaystyle A$ and $\displaystyle B$. The line through them is the perpendicular bisector.
• the bisector of an angle at $\displaystyle A$: a circle centred at $\displaystyle A$ cuts the arms at $\displaystyle P$ and $\displaystyle Q$; two circles of the same radius centred at $\displaystyle P$ and $\displaystyle Q$ meet at $\displaystyle S$. $\displaystyle \triangle APS\cong\triangle AQS$ (SSS), so $\displaystyle AS$ bisects the angle.
• the circumscribed circle: the perpendicular bisectors of two sides give the centre; the inscribed circle: the bisectors of two angles give the centre, and the perpendicular from it to a side gives the radius.

The vertices of the triangle $\displaystyle ABC$ are $\displaystyle A(0,0)$, $\displaystyle B(6,0)$ and $\displaystyle C(2,4)$.

The altitude from $\displaystyle C$ is on the line $\displaystyle {x}={2}$, and the altitude from $\displaystyle A$ is on the line $\displaystyle {y}={x}$.
Find the point $\displaystyle H$ where they meet.

Write the equation of the line of the altitude from $\displaystyle B$, and show that it passes through $\displaystyle H$ too.