In the trapezium $ABCD$, $AB // DC$ and $P$ and $Q$ are the midpoints of sides $AD$ and \(

Question

In the trapezium $ABCD$, $AB // DC$ and $P$ and $Q$ are the midpoints of sides $AD$ and $BC$ respectively. The extensions of $BP$ and $CD$ intersect at $S$. Prove that $P$ is the midpoint of $BS$, hence show that $PQ // DC$, $PQ = \frac{1}{2}(AB + DC)$.

Accepted Answer

(a) Proved: $P$ is the midpoint of $BS$ and $PQ\parallel DC$ (b) Proved: $PQ=\frac12(AB+DC)$

Proofs in Plane Geometry - Midpoint Theorem

Proofs in Plane Geometry - Midpoint Theorem

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