If AF=FB and AH=HC, then FH is the midsegment of â–³ ABC.
Always
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Let's notice that since AF=FB and AH=HC, the segment FH is the midsegment of â–³ ABC. This means that, according to the Triangle Midsegment Theorem, the length of FH is one half the length of BC.
FH=1/2BCNext, let's recall that in a trapezoid the midsegment has a length that is half of the sum of the lengths of its bases. Therefore, in trapezoid FHCB the midsegment DE has a length of half of the sum of the lengths of BC and FH.
DE=1/2(BC+FH)
Now we will substitute the value of FH and simplify.
