One Circle Touches All Three Corners, Another All Three Sides

Yes, and exactly one can — the circumcircle. A second circle, the incircle, fits inside touching all three sides. Both are found with nothing but a ruler and compasses.

The two use different constructions, and mixing them up is the usual error. This page covers everything in the ICSE Class 7 Mathematics chapter's second part: SSS, SAS and ASA constructions, special triangles, and the circumcircle and incircle.

This is the SSS construction, needing only a ruler and compasses.

To build a triangle with sides 6 cm, 5 cm and 4 cm:

1. Draw the longest side as the base, BC = 6 cm.
2. Open the compasses to 5 cm, place the point at B, and draw an arc above the base.
3. Open them to 4 cm, place the point at C, and draw a second arc crossing the first.
4. Mark the crossing point A, and join AB and AC.

The arcs cross at exactly one point above the base, so three side lengths fix one triangle.

Before starting, check the lengths are possible. The sum of any two sides must exceed the third, and with the two shortest tested against the longest:

$$4+5=9>6$$

so the triangle exists. Sides of 2 cm, 3 cm and 9 cm would fail, since $2+3=5<9$, and the arcs would never meet.

Drawing the longest side first is a practical habit rather than a rule — it keeps both arcs comfortably on the page.

And the boundary case matters: if the two shorter sides add to exactly the longest, as with 4, 5 and 9, the arcs touch on the base itself and the three points are collinear, enclosing no area at all.

Each set of measurements has its own method, and the word included decides both.

SAS — two sides and the angle between them. To build AB = 5 cm, $\angle B=60°$, BC = 6 cm:

1. Draw BC = 6 cm.
2. At B, construct a $60°$ angle — with compasses, using the equal-radius method.
3. Mark 5 cm along that arm to fix A.
4. Join AC.

ASA — two angles and the side between them. To build BC = 6 cm, $\angle B=50°$, $\angle C=70°$:

1. Draw BC = 6 cm.
2. Construct $50°$ at B and $70°$ at C.
3. Extend both arms until they meet at A.

For ASA there is a check worth doing first: the two given angles must total **less than $180°$**, or the arms will never meet. Here $50+70=120$, so the triangle exists — but angles of $100°$ and $90°$ describe nothing.

The third angle is then forced. In the ASA example it must be $180-120=60°$, which you can verify by measurement once drawn.

What makes these work is that the given part lies between the other two. Two sides with a non-included angle can fit two different triangles, which is why that combination is not a construction case at all.

Each is a special case of the methods above, with one measurement repeated or fixed.

Equilateral, side 5 cm. Draw BC = 5 cm, then draw arcs of radius 5 cm from both B and C. They cross at A, and every side is 5 cm. Each angle measures $60°$, since $\frac{180}{3}=60$.

Isosceles, base 4 cm with equal sides 6 cm. Draw BC = 4 cm, then arcs of radius 6 cm from both B and C, crossing at A. Measuring the base angles shows them equal, as the angles opposite equal sides must be.

Right-angled, legs 3 cm and 4 cm. Draw BC = 4 cm, construct a **$90°$ angle at B, mark 3 cm along it to get A, and join AC. Measuring AC gives 5 cm**, which Pythagoras confirms:

$$3^2+4^2=9+16=25=5^2$$

Isosceles right-angled. The right angle is the vertex angle, so the base angles are each $\frac{180-90}{2}=45°$.

Verifying by measurement is part of the question, not an optional extra. Construct the 3-4 triangle and measure the third side: getting 5 cm confirms both the drawing and the theorem, and getting 5.4 cm tells you the $90°$ angle was drawn carelessly.

They come from two different bisectors, and that is the distinction to hold on to.

Circumcircle — through all three vertices:

1. Construct the perpendicular bisector of any two sides.
2. They meet at the circumcentre, O.
3. With centre O and radius OA, draw the circle. It passes through B and C as well.

It works because every point on the perpendicular bisector of a side is equidistant from that side's two endpoints, so the meeting point is equidistant from all three vertices — and that common distance is the radius.

Incircle — touching all three sides:

1. Construct the bisector of any two angles.
2. They meet at the incentre, I.
3. Drop a perpendicular from I to any side to get the radius, then draw the circle.

It works because every point on an angle bisector is equidistant from the two arms, so the meeting point is equidistant from all three sides.

The positions differ, and this is examined. The incentre is always inside the triangle. The circumcentre moves: inside for an acute-angled triangle, exactly on the midpoint of the hypotenuse for a right-angled one, and outside the triangle for an obtuse-angled one.

So an obtuse triangle's circumcircle still passes through all three vertices, but its centre sits outside the shape — which surprises students who expect a centre to be enclosed by the figure.