Median Through Hypotenuse

Theorem: In a right triangle, length of a median drawn through the vertex having right angle to meet hypotenuse, is equal to one half of the length of the hypotenuse.

Prerequisites:
Median (definition)
Midpoint Theorem (proof)
SAS congruence (proof)
Angle on a straight line (proof)
Corresponding angles property (proof)

Proof:

Let $\triangle ABC$ be a right triangle, right angled at B. Let BD be a median drawn from B to meet AC at D.

We need to show that BD = $\dfrac{1}{2}$ AC. For this, let us join DE, where E is the midpoint of AB.

Since, D and E are the midpoints of AC and AB respectively (see definition of median), hence by Midpoint Theorem,

$\begin{align}\qquad\quad &DE \parallel BC\\ \Rightarrow\quad\;\; &\angle AED = \angle ABC = 90^o\quad\qquad\qquad\!\! && \text{(corresponding angles)} &&& \cdots\text{(1)}\\ \Rightarrow\quad\;\; & \angle BED = 180^o -\angle AED = 90^o && \text{(angle on a straight line)} &&& \cdots\text{(2)}\end{align}$

Now, consider $\triangle AED$ and $\triangle BED$:

$\begin{align}\qquad\quad\;\; & AE = BE\qquad\qquad\qquad\qquad\qquad && \text{(by construction)}\\ & DE = DE && \text{(common)}\\ & \angle AED = \angle BED = 90^o && \text{(from $(1)$ and $(2)$)}\\ \therefore\quad\;\;\;\;\; & \triangle AED\cong\triangle BED && \text{(by SAS congruence)}\\ \\[12pt] \Rightarrow\;\quad\quad & AD = BD && \text{(CPCTC)}\qquad\qquad\qquad\qquad\qquad\cdots\text{(3)}\\ \text{But, }\;\quad & AD = AC/2 && \text{(by definition of median)}\\ \text{Hence, }\; & BD = AC/2 && \text{(from $(3)$)}\end{align}$

Q.E.D.

Recommended;
Location of centroid
Pythagoras Theorem
Angle Bisector Theorem