Property 1 — Median to the hypotenuse

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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 \[ \angle A=180^\circ-(\angle ABC+\angle ACB) =180^\circ-2(x+y)=90^\circ. \]

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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:

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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 \( AC \).

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 \[ \angle A=90^\circ. \]