In the attached figure D and E are respectively mid points of AB and AC. Given that DE parallel to BC. Prove that DE bisects AF, that is to prove that AO = OF. I want a proof not using the concept of similarity of triangles. Please help.

Since DE parallel BC therefore OE parallel FC . Now in triangle AFC , OE parallel FC ,and E is the mid point of AC , therefore O is the mid point of AC [ Converse of mid point theorem }

Try it using sine law or you can also do it by area of triangles.See if this helps . i will post whole solution later.
EDIT:
Oh i thought we have to prove mid point theorem.
You can prove it by converse of mid point theorem

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TopNewestSince DE parallel BC therefore OE parallel FC . Now in triangle AFC , OE parallel FC ,and E is the mid point of AC , therefore O is the mid point of AC [ Converse of mid point theorem }

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In Triangle ABF D mid point of AB and DO//BF so O is mid point of AF

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Try it using sine law or you can also do it by area of triangles.See if this helps . i will post whole solution later. EDIT: Oh i thought we have to prove mid point theorem. You can prove it by converse of mid point theorem

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