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

Beginner2023-02-20Added 9 answers

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

$\mathrm{\angle}DEC=\mathrm{\angle}BEF\phantom{\rule{0ex}{0ex}}CE=EB\phantom{\rule{0ex}{0ex}}\mathrm{\angle}DCE=\mathrm{\angle}EBF\phantom{\rule{0ex}{0ex}}\therefore \mathrm{\u25b3}CDE\cong \mathrm{\u25b3}BFE\phantom{\rule{0ex}{0ex}}\therefore DC=BF\phantom{\rule{0ex}{0ex}}AB=BF\phantom{\rule{0ex}{0ex}}AF=AB+BF=AB+AB\phantom{\rule{0ex}{0ex}}=2AB\phantom{\rule{0ex}{0ex}}AF=2AB$

Proof : In CDE and EBF

$\mathrm{\angle}DEC=\mathrm{\angle}BEF\phantom{\rule{0ex}{0ex}}CE=EB\phantom{\rule{0ex}{0ex}}\mathrm{\angle}DCE=\mathrm{\angle}EBF\phantom{\rule{0ex}{0ex}}\therefore \mathrm{\u25b3}CDE\cong \mathrm{\u25b3}BFE\phantom{\rule{0ex}{0ex}}\therefore DC=BF\phantom{\rule{0ex}{0ex}}AB=BF\phantom{\rule{0ex}{0ex}}AF=AB+BF=AB+AB\phantom{\rule{0ex}{0ex}}=2AB\phantom{\rule{0ex}{0ex}}AF=2AB$