Given in triangle ABC, a straight line AG is drawn through the vertex A .E and F are respectively mid points of sides AB and AC .It is joined intersecting the line AG at O.Now ,in triangle ABC ,EF =1/2BC and EF||BC (by mid point theorem) (1).Now in triangle AGC ,F is mid point and OF is parallel to GC from equation 1.hence O is mid point of AG( by converse of mid point theorem.Hence AG is biscte by EF
						

							Please follow the solution to the above given query:GIVEN :   ​ABC is a triangle, in which D is the mid - point of AB and E is the mid-point of AC.             AL is a straight line that intersects DE at M and BC at L.  TO PROVE : DE bisects AL.PROOF : Since D and E are the mid points of the sides AB and AC of  △ ABC, then                   DE || BC      .....(1)     [Mid point theorem]Since DE || BC, then DM || BL.   [as DM is a part of DE and BL is a part of BC].In  triangle ABL, we have D is the mid point of AB and DM || BL , thenM must be the mid point of AL.   [Converse of mid point theorem]⇒ AM = ML⇒ DE bisects AL Hope this information will clear your doubts about Triangles.