Know these cold: median to the hypotenuse \(=\) half the hypotenuse (circumradius); equal segments converse \(\Rightarrow\) right angle; and that hypotenuse-median is the shortest.
Property 1 — Median to the hypotenuse
In a right triangle \( \triangle ABC \) with right angle at \( B \), let \( D \) be the midpoint of the hypotenuse \( AC \). Then
\[
BD=AD=CD=\frac{AC}{2}.
\]
The midpoint of the hypotenuse is equidistant from \(A,B,C\); hence the circle with center \(D\) and radius \( \tfrac{AC}{2} \) passes through \(A,B,C\).
Show figureRight triangle with \( \angle B=90^\circ \) and \( D \) midpoint of \( AC \).
Property 2 — A right-angle converse from equal segments
Let \( D \) be the midpoint of \( AC \) in any triangle \( \triangle ABC \). If
\[
BD = AD = DC,
\]
then \( \angle A = 90^\circ \).
Complete proof
From \( AD=BD \) we get \( \triangle ABD \) isosceles, so \( \angle ABD=\angle ADB=x \).
From \( BD=DC \) we get \( \triangle BDC \) isosceles, so \( \angle BCD=\angle CBD=y \).
Now at vertex \( B \): \( \angle ABC=\angle ABD+\angle DBC=x+y \).
At vertex \( C \): \( \angle ACB=\angle ACD+\angle DCB=y+x \).
Hence
\[
\angle A=180^\circ-(\angle ABC+\angle ACB)
=180^\circ-2(x+y)=90^\circ.
\]
Show figureEqual segments force complementary base angles: \( x+y=90^\circ \).
Property 3 — In a right triangle, that median is the shortest
In \( \triangle ABC \) right-angled at \( B \), denote medians by \( m_a,m_b,m_c \) from sides \( a=BC,\ b=CA,\ c=AB \) respectively. Then the median to the hypotenuse is the smallest:
Show figureHere \( m_b=BD=\tfrac{AC}{2} \) is the smallest median.
CAT-style Question 1
Question. In a right-angled triangle \( \triangle ABC \) with \( \angle B=90^\circ \) and \( AC=10\,\text{cm} \), find the length of the median from \( B \) to \( AC \).
Solution
By Property 1, the median to the hypotenuse equals half the hypotenuse:
\[
BD=\frac{AC}{2}=\boxed{5\,\text{cm}}.
\]
CAT-style Question 2 (converse)
Question. In triangle \( \triangle ABC \), let \( D \) be the midpoint of \( AC \). If \( AD=BD=DC \), determine \( \angle A \).
Solution (with figure on click)
As proved in Property 2, equal segments force \( x+y=90^\circ \), hence
\[
\angle A=90^\circ.
\]
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