Problem — A median equal to half a side implies a right triangle

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In a triangle $RST$ the median to the side $RS$ has length equal to half of $RS$ itself. Letting $M$ be the midpoint of $RS$, we thus have $TM \cong \dfrac{1}{2}\,RS$. Prove that the triangle is right-angled, with the right angle at the vertex $T$ opposite the side $RS$.

Solution

Since $M$ is the midpoint of $RS$, we have $RM \cong MS \cong \dfrac{1}{2}\,RS$. By hypothesis the median $TM$ also equals half of $RS$, hence $RM \cong MS \cong TM.$

Triangle $RMT$ has $RM \cong TM$: it is isosceles with base $RT$, so its base angles are congruent, $\widehat{MRT} \cong \widehat{MTR} = x.$ Likewise triangle $SMT$ has $MS \cong TM$: it is isosceles with base $ST$, so $\widehat{MST} \cong \widehat{MTS} = y.$

The angles of triangle $RST$ are then $\widehat{R} = x$, $\widehat{S} = y$ and $\widehat{T} = \widehat{MTR} + \widehat{MTS} = x + y$. From the sum of the interior angles: $x + y + (x + y) = 180^\circ \quad\Longrightarrow\quad 2(x + y) = 180^\circ \quad\Longrightarrow\quad x + y = 90^\circ.$ But $\widehat{T} = x + y$, hence $\widehat{T} = 90^\circ$: the triangle is right-angled with the right angle at $T$, opposite the side $RS$. $\;\blacksquare$