Suppose abcd is a rhombus and that the bisector of ? abd meets ad at point k. prove that m? akb = 3m? abk.

Given: ae ≅ ce ; de ≅ be prove: abcd is a parallelogram. we have that ab || dc. by a similar argument used to prove that △aeb ≅ △ced, we can show that △ ≅ △ceb by. so, ∠cad ≅ ∠ by cpctc. therefore, ad || bc by the converse of the theorem. since both pair of opposite sides are parallel, quadrilateral abcd is a parallelogram.