Median, altitude and angle bisector: three different tasks

We need to mark a triangular part “down the middle”. The middle of what: its bottom edge, its top angle or its perpendicular height? These are three different requirements. They produce one line only under additional conditions.

A median joins a vertex to the midpoint of the opposite side; an altitude runs perpendicularly from a vertex to the line of the opposite side; an angle bisector halves the angle at a vertex. The prerequisites are triangle elements and congruence criteria.

Three precise requirements

Draw the segments from A in triangle ABC. For median AM, first find midpoint M of BC. Its defining condition is:

$BM=MC.$

Join A to M. No right angle is required, and angle A need not be halved.

For altitude AH, draw the perpendicular from A to line BC, meeting it at foot H. The requirement is AH⊥BC, not BH=HC. For angle bisector AL, take the internal ray halving angle A and stop it at side BC at L:

$\angle BAL=\angle LAC.$

An angle bisector as a ray is unbounded; the triangle’s bisector here is this segment. Equal angles do not guarantee BL=LC.

Different tasks, different segments

AM · median · solid

AH · altitude · dashed

AL · bisector · dotted

AB = AC: one axis from vertex A

An altitude to an extended side

Arbitrary units, the same uniform scale. The first triangle is scalene. Distinct H, L, M enforce a right angle, equal angles and equal base parts respectively. Second: equal lateral sides give AM=AH=AL as one segment. Third: H is outside BC; an altitude reaches line BC, not necessarily an interior point of the side.

H, L and M are distinct in the first diagram. Different colours and line patterns identify the segments; sharing vertex A does not make them the same line.

An altitude can lie outside a side

An altitude reaches the line containing the side. In an obtuse triangle two altitude feet lie on extended sides, so those altitudes run outside the interior. In a right triangle the two legs are already perpendicular, so either is an altitude to the other. We need not force all three altitudes inside.

Every vertex has a median, an altitude and an internal bisector: three segments of each kind. In a scalene triangle we cannot assume that they coincide.

Why an isosceles triangle has one axis from its apex

Suppose AB=AC and M is the midpoint of BC. In triangles ABM and ACM, AB=AC, BM=MC and AM is shared. SSS makes the triangles congruent.

Hence angles BAM and MAC are equal: AM is an angle bisector. Angles AMB and AMC are also equal. They form a linear pair summing to 180°, so each is 90°: AM is an altitude. Starting with the median, we proved the other properties.

Thus, from the vertex between the equal sides, the median, altitude and angle bisector coincide. Their line is a symmetry axis: one half of the part reflects onto the other. The base angles are equal; in an equilateral triangle all angles are 60° and an axis runs from every vertex. Euclid I.5.

For symmetric marking, first check AB=AC, then halve the base. Without equal lateral sides, the base midpoint guarantees neither a perpendicular nor a halved apex angle.

Additional factOptional · you may skip this

Three families, different intersection points

The three medians meet at the centroid, dividing each in the ratio 2:1 from the vertex. This is a separate theorem; the university explanation proves it through similarity or coordinates, rather than a drawing’s appearance.

The three altitude lines meet at the orthocentre, and the three internal angle bisectors at the incentre. These too are separate facts, with proofs in the AMSI material. In a general triangle the points need not coincide. In an equilateral triangle symmetry makes them coincide. The main explanation above does not require these facts.