Question

In triangle ABC, D is midpoint of BC, and AD is extended to G such that OD = DG. CE and BF intersect at O on AD. (i) In quadrilateral OBGC, diagonals OG and BC bisect each other at D, hence OBGC is a parallelogram. (ii) Since OBGC is parallelogram, CO ∥ GB, thus CE ∥ GB. In triangle AGB, OE ∥ GB implies by BPT: $\frac{AO}{OG}=\frac{AE}{EB}$. Similarly in triangle AGC, OF ∥ GC gives $\frac{AO}{OG}=\frac{AF}{FC}$. Hence $\frac{AE}{EB}=\frac{AF}{FC}$, so EF ∥ BC by converse of BPT. (iii) In triangles AEF and ABC, $\angle AEF=\angle ABC$ (corresponding angles since EF ∥ BC) and $\angle A$ is common, so $△AEF\sim △ABC$ by AA similarity.