In parallelogram $ABCD$, $E$ is the midpoint of side $AD$. Line $BE$ meets line $CD$ extended at $F$. Then…

In parallelogram $ABCD$, $E$ is the midpoint of side $AD$. Line $BE$ meets line $CD$ extended at $F$. Then $DF$ equals:
  1. $\dfrac{1}{2} CD$
  2. $CD$
  3. $2 \cdot CD$
  4. $\dfrac{3}{2} CD$

Solution

Triangles $BEA$ and $FED$ are congruent (vertically opposite angles at $E$, $AE = ED$, alternate angles since $AB \parallel DF$). Hence $DF = AB = CD$.

Asked in: IMO

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