$A B + C D = 2 E F$
If the above statement is true then mention answer as 1, else mention 0 if false

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Solution

In $△ F D P$ and $△ A D B$ ,
$∠ A D B = ∠ F D P$ (Common angle)
$∠ D A B = ∠ D F P$ (Corresponding angles)
$∠ D B A = ∠ D P F$ (Corresponding angles)
Thus, $△ F D P \sim △ A D B$ (AAA rule)
hence, $\frac{F P}{A B} = \frac{F D}{A D}$
$\frac{F P}{A B} = \frac{1}{2}$ (Since, AF = FD)
or $F P = \frac{1}{2} A B$ ...(I)

Since, $A B ∥ F B ∥ C D$
By equal Intercept theorem, $B E = E C$
Hence, $P E = \frac{1}{2} C D$ ...(II)
Thus, adding I and II
$P F + P E = \frac{1}{2} (A B + C D)$
$2 E F = A B + C D$

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