Problem — In a trapezoid the difference of the bases is less than the sum of the legs

Text

Let $PQRS$ be a trapezoid with bases $PQ$ and $SR$, where $PQ \parallel SR$ and $PQ > SR$; the legs are $SP$ and $QR$. Prove that the difference of the bases is less than the sum of the legs, that is $PQ - SR < SP + QR.$

(original figure: trapezoid $PQRS$ with the parallel to the leg $SP$ drawn through $R$)

Solution

Going around the boundary, the vertices are $P$ and $Q$ on the longer base and $R$, $S$ on the shorter base, with $SR \parallel PQ$. Through $R$ draw the parallel to the leg $SP$; it meets the base $PQ$ at a point $T$. The quadrilateral $SPTR$ has $SR \parallel PT$ (they lie on the two bases) and $SP \parallel RT$ (by construction), hence it is a parallelogram. Therefore $PT \cong SR,\qquad RT \cong SP.$ Since $PT \cong SR < PQ$, the point $T$ falls inside the segment $PQ$, and so $TQ = PQ - PT = PQ - SR.$ Consider the triangle $TQR$ (non-degenerate, since $R$ does not lie on the line $PQ$). By the triangle inequality one side is less than the sum of the other two: $TQ < QR + RT.$ Substituting $TQ = PQ - SR$ and $RT \cong SP$ we obtain $PQ - SR < QR + SP.$ $\boxed{\,PQ - SR < SP + QR\,}$ $\blacksquare$