Alan Wright

2023-02-19

In the figure, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove that AF = 2AB.

Roy Mcclain

Given: ABCD in ||gm, with E as BC's midpoint. To meet AB produced at F, DE is joined and created.
Proof : In CDE and EBF
$\mathrm{\angle }DEC=\mathrm{\angle }BEF\phantom{\rule{0ex}{0ex}}CE=EB\phantom{\rule{0ex}{0ex}}\mathrm{\angle }DCE=\mathrm{\angle }EBF\phantom{\rule{0ex}{0ex}}\therefore \mathrm{△}CDE\cong \mathrm{△}BFE\phantom{\rule{0ex}{0ex}}\therefore DC=BF\phantom{\rule{0ex}{0ex}}AB=BF\phantom{\rule{0ex}{0ex}}AF=AB+BF=AB+AB\phantom{\rule{0ex}{0ex}}=2AB\phantom{\rule{0ex}{0ex}}AF=2AB$

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