The median to the hypotenuse of a right triangle

A median of a triangle is a segment from a vertex to the midpoint of the side facing it.

• In a right triangle, the side facing the right angle is the hypotenuse.
• The median from the vertex of the right angle goes to the midpoint of the hypotenuse: it is the median to the hypotenuse.

Theorem: in a right triangle, the median to the hypotenuse is half the hypotenuse.

• In triangle $\displaystyle ABC$, angle $\displaystyle B$ is right, and $\displaystyle M$ is the midpoint of the hypotenuse $\displaystyle AC$.
• Then $\displaystyle {BM}={\frac{1}{2}AC}$, that is, $\displaystyle {BM}={AM}={CM}$.

Given: $\displaystyle {\angle ABC}={90^{\circ}}$, and $\displaystyle BM$ is the median to the hypotenuse.
Prove: $\displaystyle {BM}={\frac{1}{2}AC}$.

The proof below extends the median $\displaystyle BM$ by its own length, to the point $\displaystyle D$, making a quadrilateral $\displaystyle ABCD$:

• its diagonals bisect each other, so it is a parallelogram;
• it has a right angle, so it is a rectangle;
• a rectangle’s diagonals are equal, and $\displaystyle BM$ is half a diagonal.

Converse theorem: if a median of a triangle is half the side it goes to, then the angle at the vertex it comes from is right.

Given: $\displaystyle BM$ is a median of triangle $\displaystyle ABC$, and $\displaystyle {BM}={\frac{1}{2}AC}$.
Prove: $\displaystyle {\angle ABC}={90^{\circ}}$.

The proof below uses the fact that the median splits the triangle into two isosceles triangles, whose base angles make up all the angles of the triangle:

In triangle $\displaystyle ABC$, angle $\displaystyle B$ is right, $\displaystyle {AB}={6}$, $\displaystyle {BC}={8}$, and $\displaystyle BM$ is the median to the hypotenuse.

• By Pythagoras: $\displaystyle {AC}={\sqrt{6^2+8^2}}={\sqrt{100}}={10}$;
• so $\displaystyle {BM}={\frac{1}{2}\cdot10}={5}$, and also $\displaystyle {AM}={CM}={5}$;
• triangle $\displaystyle ABM$ is isosceles ($\displaystyle {AM}={BM}$), so $\displaystyle {\angle ABM}={\angle A}$.

Calculate, and give the reason for every step.

The legs of a right triangle are $\displaystyle 5$ and $\displaystyle 12$ long.
How long is the median to the hypotenuse?

In triangle $\displaystyle KLM$, angle $\displaystyle L$ is right.
The median to the hypotenuse, $\displaystyle LN$, is $\displaystyle 7.5$ long, and $\displaystyle {KL}={9}$.
Find $\displaystyle LM$.

In triangle $\displaystyle ABC$, angle $\displaystyle B$ is right, and $\displaystyle BM$ is the median to the hypotenuse.
Angle $\displaystyle A$ is $\displaystyle 28^{\circ}$.
Find $\displaystyle \angle ABM$ and $\displaystyle \angle BMC$.

In triangle $\displaystyle PQR$, the segment $\displaystyle PS$ is the median to the side $\displaystyle QR$.
$\displaystyle {PS}={5}$ and $\displaystyle {QR}={10}$.
How many degrees is angle $\displaystyle QPR$?

In quadrilateral $\displaystyle ABCD$, the angles $\displaystyle B$ and $\displaystyle D$ are right, and $\displaystyle M$ is the midpoint of the diagonal $\displaystyle AC$.
Prove: $\displaystyle {MB}={MD}$.

$\displaystyle ABCD$ is a rhombus.
Its diagonals meet at the point $\displaystyle O$, and $\displaystyle K$ is the midpoint of the side $\displaystyle AB$.
Prove: $\displaystyle {OK}={\frac{1}{2}AB}$.

In quadrilateral $\displaystyle ABCD$, the point $\displaystyle M$ is the midpoint of the diagonal $\displaystyle AC$, and $\displaystyle {MB}={MD}={MA}$.
Prove: $\displaystyle {\angle ABC}={\angle ADC}$.