Question

IN THE GIVEN FIGURE AD, CE ARE THE MEDIANS OF TRIANGLE ABC, DF IS DRAWN PARALLEL TO CE . PROVE THAT ,EF=FB, AG:GD =2:1​

Answer 1

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Yeah, I solved it! it's quite simple now. As we know, AD is the median. so, D is the midpoint of BC Therefore, BD = DC.........(1) Therefore, BD/CD= 1:1.........(2) now in triangle ABC, DF is parallel to BE . Therefore, CF/EF=CD/BD,........... By BPT (basic proportionality theorem) .............(3). Therefore from equation (2), CF/EF = 1:1 Therefore, CF is equal to EF Therefore EF=FC. now as given in the figure AD and BE are medians They intersect at the G. Thus the point of intersection of two medians is G. Does G is centroid here. We know that a centroid divides any mediun in the proportion 2:1.......(4). Therefore, AG:GD= 2:1 This question is solved hope you will understand.