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Every triangle has four families of special segments, and it's easy to mix them up. Here's a clean reference:

Median — connects a vertex to the midpoint of the opposite side. All three meet at the centroid.
Altitude — perpendicular from a vertex to the opposite side (or its extension). All three meet at the orthocenter.
Angle bisector — splits a vertex angle in half. All three meet at the incenter.
Midsegment — connects the midpoints of two sides; parallel to the third side and half its length.

Key formulas

Median length (from the vertex opposite side $a$ , using the other two sides $b, c$ ):

m_a = \dfrac12\sqrt{2b^2+2c^2-a^2}

Altitude, two ways:

h_a = \dfrac{2\cdot\text{Area}}{a} \qquad\text{(general)} \qquad h = \dfrac{\text{leg}_1 \cdot \text{leg}_2}{\text{hypotenuse}} \qquad\text{(right triangle only)}

Angle bisector length (from the vertex between sides $b, c$ , opposite angle $A$ ):

t_a = \dfrac{2bc}{b+c}\cos\!\left(\dfrac{A}{2}\right)

A fact worth remembering even without the formula: the three medians divide a triangle into six smaller triangles of equal area.

Worked example

Problem: Triangle $ABC$ has $AB=13$ , $AC=15$ , and $BC=14$ . Find the length of the median from $A$ to the midpoint of $BC$ .

Solution: This median is drawn to side $a=BC=14$ , using the other two sides $b=AC=15$ and $c=AB=13$ :

m_a = \dfrac12\sqrt{2(15)^2+2(13)^2-14^2} = \dfrac12\sqrt{450+338-196} = \dfrac12\sqrt{592}

= \dfrac12 \cdot 4\sqrt{37} = 2\sqrt{37}

1.A right triangle has legs 6 and 8. Find the altitude drawn to the hypotenuse.

2.The three medians of a triangle divide it into how many smaller triangles of equal area, and what fraction of the total area is each?

3.A triangle has sides $a=8, b=10, c=12$ . Find the length of the median to side $a$ .

4.In a triangle with area 60 and side $BC=15$ , find the altitude from $A$ to $BC$ .