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In triangle ABC, let D be on AB and E be on AC such that DE is parallel to BC. Let DC and BE intersect at H and AH extended intersects BC at F. Prove F is the midpoint of BC.
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do you have a drawing?
no
ok got a drawing
|dw:1415237488527:dw|
and DE is parallel to BC
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ok let me think
may i add that bcde is a trapezoid, with dc and be as diagonals
that is true
maybe i could prove af is a median?
yeah that's essentially what they want, but I'm trying to remember how to do it. hmm I remember seeing a proof for it
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all i know right know is that triangles dhe and chb
are similar
that is correct
since these pairs of angles are congruent |dw:1415238195666:dw|
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