Triangle Medians — 3 Properties You Must Know | AzuCATion
GeometryTrianglesCAT · XAT4–6 minEasy–Moderate
Triangle Medians — 3 Properties You Must Know
A median joins a vertex to the midpoint of the opposite side. These three facts appear again and again in CAT/XAT geometry. Learn the idea + a 10-second proof each.
Property 1 — Median to the hypotenuse
Statement. In a right triangle \(\triangle ABC\) with right angle at \(B\), let \(D\) be the midpoint of the hypotenuse \(AC\). Then
\[
BD=AD=CD=\frac{AC}{2}.
\]
Moreover, the circle with center \(D\) and radius \(\tfrac{AC}{2}\) passes through \(A,B,C\) (Thale’s circle).
Why. In a right triangle, the midpoint of the hypotenuse is equidistant from all three vertices (via congruent triangles / perpendicular-bisector logic).
Show figureWith \(\angle B=90^\circ\) and \(D\) midpoint of \(AC\), we get \(BD=AD=CD=AC/2\).
Property 2 — A right-angle converse from equal segments
Statement. In any triangle \(\triangle ABC\), if a point \(D\) on \(AC\) satisfies \(BD=AD=DC\), then \(\angle A=90^\circ\).
Proof in one line: From \(AD=BD\) we get isosceles \(\triangle ABD\Rightarrow \angle ABD=\angle ADB=x\).
From \(BD=DC\) we get isosceles \(\triangle BDC\Rightarrow \angle CBD=\angle BCD=y\).
Then \(\angle ABC=x+y\) and \(\angle ACB=y+x\). Hence
\[
\angle A=180^\circ-(\angle ABC+\angle ACB)
=180^\circ-2(x+y)=90^\circ.
\]
Show figureEqual segments force complementary base angles; thus the included angle at \(A\) is right.
Property 3 — In a right triangle, that median is the shortest
In \(\triangle ABC\) right-angled at \(B\), let medians to sides \(a=BC\), \(b=CA\) (hypotenuse), \(c=AB\) be \(m_a, m_b, m_c\) respectively. Then
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