



 
Carolyn, The key is in similarities, in particular similar triangles. Start with a triangle ABC and a point D on BC so that the ratio of the lengths of BD to DC is p to q, in your case take p = 1 and q = 2. Construct a line through D parallel to CA and meeting AB at E. Construct a line through D parallel to AB and meeting CA at F. AEDF is a parallelogram and hence AE and DF are congruent. The triangles CFD and DEB are similar (do you see why?) and hence
Does this help?  


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