Answer:
ΔABC has: AD=DB
AE = EC
=> DE is the median line of ΔABC
=> DE // BC
=> m∠AED = m∠ECF = 32°
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Step-by-step explanation:
∠ ECF = 32°
DE is the midsegment of Δ ABC and is therefore parallel to the third side BC.
Since DE and BC are parallel lines, then
∠ ECF and ∠ AED are congruent ( corresponding angles ) , thus
∠ ECF = ∠ AED = 32°
