Problem — A triangle divided by its midpoints into four congruent triangles

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In a triangle $RST$, let $U$, $V$ and $W$ be the midpoints of the sides $ST$, $TR$ and $RS$ respectively. Prove that the three segments $UV$, $VW$ and $WU$ divide the triangle into four congruent triangles.

(original figure: triangle $RST$ with the midpoints of the sides and the resulting subdivision into four triangles)

Solution

The three segments split the triangle into the four triangles $RWV$, $WSU$, $VUT$ and the central triangle $UVW$. By the midsegment theorem, the segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it. Applying it to the three sides of $RST$:

• $VW \cong \tfrac12 ST$;
• $WU \cong \tfrac12 TR$;
• $UV \cong \tfrac12 RS$.

Moreover, since $U,V,W$ are midpoints of the sides, each half-side is half of the whole side: $RW \cong WS \cong \tfrac12 RS,\qquad TV \cong VR \cong \tfrac12 TR,\qquad SU \cong UT \cong \tfrac12 ST.$ Let us then list the sides of the four triangles:

• $RWV$: sides $RW \cong \tfrac12 RS$, $RV \cong \tfrac12 TR$, $WV \cong \tfrac12 ST$;
• $WSU$: sides $WS \cong \tfrac12 RS$, $SU \cong \tfrac12 ST$, $WU \cong \tfrac12 TR$;
• $VUT$: sides $VT \cong \tfrac12 TR$, $TU \cong \tfrac12 ST$, $UV \cong \tfrac12 RS$;
• $UVW$: sides $UV \cong \tfrac12 RS$, $VW \cong \tfrac12 ST$, $WU \cong \tfrac12 TR$.

Each of the four triangles therefore has its three sides congruent to $\tfrac12 RS$, $\tfrac12 ST$ and $\tfrac12 TR$ respectively. By the SSS congruence criterion the four triangles are congruent to one another. $\;\blacksquare$