Triangle Rules in the Achievement Test

Triangle Rules in the Achievement Test — Moosa Academy

Triangles carry more test questions than any other shape. Almost all of them come down to four things: the angle rules, the three similarity criteria, the four special segments, and the centroid. This sheet collects them in the order they are usually needed.

Concept Angles and sides

1. The interior angles add to

180^\circ

2. Two equal sides

\leftrightarrow

two equal base angles

3. Three equal sides

\leftrightarrow

all angles

60^\circ

4. Exterior

+

adjacent interior

= 180^\circ

5. Exterior

=

sum of the two remote interior angles

6. An exterior angle is larger than either remote interior angle

7. The midsegment is half the base and parallel to it

8. In a right triangle,

a^{2} + b^{2} = c^{2}

9. The largest angle faces the longest side

10. Any point on the perpendicular bisector is equidistant from the two endpoints

11. Any point on an angle bisector is equidistant from the two arms

Rules 2 and 3 work in both directions — that is what the double arrow means. Being told the angles are equal lets you conclude the sides are, and the reverse holds too.

Theorem The three similarity criteria

Two triangles are similar when any one of AA, SAS or SSS is satisfied.

Criterion	What you need
AA	Two pairs of equal angles
SAS	Two pairs of proportional sides with equal included angles
SSS	All three pairs of sides proportional

AA needs only two angles because the third follows from the $180^\circ$ rule. For SAS the equal angle must sit between the two proportional sides — an angle elsewhere proves nothing.

Concept The four special segments

Angle bisector — splits an angle in two and reaches the opposite side

Perpendicular bisector — meets a side at its midpoint at

90^\circ

Altitude — drops perpendicularly from a vertex to the opposite side or its extension

Median — joins a vertex to the midpoint of the opposite side

The altitude and the median both run from a vertex to the opposite side, which is why they get confused. The altitude cares only about the right angle; the median cares only about the midpoint. In an isosceles triangle drawn from the apex they happen to coincide — but only there.

Theorem The centroid

The three medians meet at one point, the centroid, which sits two-thirds of the way along each median from the vertex.

Distance from the vertex:

\dfrac{2}{3}

of the median's length

Distance from the midpoint of the base:

\dfrac{1}{3}

of the median's length

So the centroid divides every median in the ratio $2 : 1$ , with the longer part nearer the vertex. Questions usually give you one of the two pieces and ask for the other or for the whole.

Example Using the exterior angle rule

In a triangle the two remote interior angles are $55^\circ$ and $63^\circ$ . Find the exterior angle at the third vertex, and that vertex's interior angle.

Exterior angle

= 55^\circ + 63^\circ = 118^\circ

Interior angle

= 180^\circ - 118^\circ = 62^\circ

Check:

55^\circ + 63^\circ + 62^\circ = 180^\circ

Exterior $118^\circ$ , interior $62^\circ$

Adding the two remote angles is faster than finding the third angle first and subtracting. The check at the end costs nothing and catches slips.

Summary

Interior angles total $180^\circ$ ; an exterior angle equals the sum of the two remote interior angles.
Equal sides and equal base angles imply each other, in both directions.
The largest angle always faces the longest side.
Similarity needs AA, SAS or SSS — and in SAS the angle must be the included one.
The four special segments are the angle bisector, perpendicular bisector, altitude and median.
An altitude is defined by a right angle; a median by a midpoint. They are not the same segment.
The medians meet at the centroid, which divides each of them in the ratio $2 : 1$ from the vertex.