The median, the altitude and the angle bisector

From every vertex of a triangle three special segments go to the side opposite it:

• the median – reaches the midpoint of the side;
• the altitude – is perpendicular to the side;
• the angle bisector – cuts the vertex’s angle into two equal angles.

A median of a triangle is a segment joining a vertex to the midpoint of the side opposite it.

• In triangle $\displaystyle ABC$ the point $\displaystyle D$ is the midpoint of $\displaystyle BC$: $\displaystyle {BD}={DC}$.
• $\displaystyle AD$ is the median to the side $\displaystyle BC$.
• Every triangle has three medians – one to each side.

An altitude of a triangle is a segment from a vertex, perpendicular to the side opposite it.

• $\displaystyle AH$ is perpendicular to $\displaystyle BC$: $\displaystyle AH$ is the altitude to the side $\displaystyle BC$.
• Every triangle has three altitudes – one to each side.
• In a right triangle, each leg is also the altitude to the other leg.

In an obtuse triangle, the altitude from the vertex of an acute angle lies outside the triangle: it is perpendicular to the extension of the side opposite it.

An angle bisector of a triangle is a segment from a vertex that cuts its angle into two equal angles and reaches the side opposite it.

• $\displaystyle {\angle BAE}={\angle EAC}$: $\displaystyle AE$ is the bisector of angle $\displaystyle A$.
• Every triangle has three angle bisectors – one for each angle.

In this triangle the three segments from $\displaystyle A$ differ from each other: the altitude $\displaystyle AH$, the angle bisector $\displaystyle AE$ and the median $\displaystyle AD$.

Draw on the board the median to the side $\displaystyle AC$, and the altitude to the side $\displaystyle AB$.

Calculate, and give the reasons.

In triangle $\displaystyle ABC$, the segment $\displaystyle AD$ is the median to the side $\displaystyle BC$, and $\displaystyle {BC}={14}$ cm.
Find the length of $\displaystyle BD$.

In triangle $\displaystyle ABC$, the segment $\displaystyle AD$ is the altitude to the side $\displaystyle BC$, and $\displaystyle {\angle B}={35^{\circ}}$.
Find the angle $\displaystyle BAD$.

In triangle $\displaystyle ABC$, the segment $\displaystyle AD$ bisects angle $\displaystyle A$.
$\displaystyle {\angle BAC}={70^{\circ}}$ and $\displaystyle {\angle B}={60^{\circ}}$.
Find the angle $\displaystyle BAD$ and the angle $\displaystyle ADB$.

The three medians of a triangle meet at one point, $\displaystyle G$ – in every triangle.

The point $\displaystyle G$ is the triangle’s centre of gravity: a triangle cut out of cardboard balances on the tip of a pencil right under $\displaystyle G$.

A median cuts the triangle into two triangles of equal area: here $\displaystyle ABD$ and $\displaystyle ADC$.

• The area of a triangle is a side times the altitude to it, divided by 2.
• The bases are equal: $\displaystyle {BD}={DC}$.
• Both triangles have the same altitude: $\displaystyle AH$, the altitude from $\displaystyle A$ to the line $\displaystyle BC$.
• So the areas are equal: each is half the area of triangle $\displaystyle ABC$.

In triangle $\displaystyle ABC$, the segment $\displaystyle AD$ is the median to the side $\displaystyle BC$.

The area of triangle $\displaystyle ABC$ is 48 cm².
What is the area of triangle $\displaystyle ABD$?

$\displaystyle {BC}={12}$ cm, and the altitude from $\displaystyle A$ to the side $\displaystyle BC$ is 5 cm.
What is the area of triangle $\displaystyle ADC$?