Question

In trapezium $ABCD$, sides $AB$ and $DC$ are parallel to each other. $F$ is mid-point of $AD$ and $E$ is mid-point of $BC$. Can it be concluded that $AB+DC=2EF$ ?
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

In $△FDP$ and $△ADB$,
$\angle ADB=\angle FDP$ (Common angle)
$\angle DAB=\angle DFP$ (Corresponding angles)
$\angle DBA=\angle DPF$ (Corresponding angles)
Thus, $△FDP\sim △ADB$ (AAA rule)
hence, $\frac{FP}{AB}=\frac{FD}{AD}$
$\frac{FP}{AB}=\frac{1}{2}$ (Since, AF = FD)
or $FP=\frac{1}{2}AB$...(I)
Since, $AB\parallel FB\parallel CD$
By equal Intercept theorem, $BE=EC$
Hence, $PE=\frac{1}{2}CD$ ...(II)
Thus, adding I and II
$PF+PE=\frac{1}{2}(AB+CD)$
$2EF=AB+CD$