Problem — Equally spaced points on an equilateral triangle

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Let $PQR$ be an equilateral triangle. On its sides, going around in the same direction, take the points $D \in PQ$, $E \in QR$ and $F \in RP$ such that $PD \cong QE \cong RF$. Prove that the triangle $DEF$ is equilateral.

Solution

Since $PQR$ is equilateral, its sides are congruent and its angles all measure $60^\circ$: $PQ \cong QR \cong RP, \qquad \widehat{P} \cong \widehat{Q} \cong \widehat{R} = 60^\circ.$

By hypothesis $PD \cong QE \cong RF$. Subtracting these congruent segments from congruent sides gives congruent remaining segments: $DQ = PQ - PD, \quad ER = QR - QE, \quad FP = RP - RF \;\Rightarrow\; DQ \cong ER \cong FP.$

Compare the triangles $DQE$, $ERF$ and $FPD$:

• $DQ \cong ER \cong FP$ (just shown);
• $QE \cong RF \cong PD$ (by hypothesis);
• the included angles $\widehat{Q} \cong \widehat{R} \cong \widehat{P}$ are congruent (each is $60^\circ$).

By the first congruence criterion (SAS) the three triangles are congruent, so their third sides are congruent: $DE \cong EF \cong FD.$

The triangle $DEF$ has three congruent sides: it is equilateral. $\blacksquare$