l have attached a picture.
In that figure , BA || DF and CA || EG and BD=EC. Prove that
1. BG=DF
2. EG=CF
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in ∆ABC,
BD=EC(given) ...(i)
therefore, BD+DE=EC+DE
thus,BE=DC
in ∆ BEG & ∆ DFC
<GBE= <FDC (BA||DF)
BE=EC( from (i))
<GEB=<FCD(CA||EG)
therefore ∆BEG & ∆CDF are congruent to each other .
thus,
BG=DF(cpct*)
EF=CF(cpct)
*corresponding part of a congruent triangle
BD=EC(given) ...(i)
therefore, BD+DE=EC+DE
thus,BE=DC
in ∆ BEG & ∆ DFC
<GBE= <FDC (BA||DF)
BE=EC( from (i))
<GEB=<FCD(CA||EG)
therefore ∆BEG & ∆CDF are congruent to each other .
thus,
BG=DF(cpct*)
EF=CF(cpct)
*corresponding part of a congruent triangle
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