Question

cho ΔDEF cân tại D, C là trung điểm EF

a)CMR: ΔDCE=ΔDLF

b)CM: DC⊥EF

c)kẻ CA⊥DE; CB⊥DF. CMR ΔDAC=ΔDBC

Answer 1

a) Xet △DCE va △DCF co:

\(\left\{{}\begin{matrix}DE=DF\left(tgDEFcan\right)\\chungDC\\EC=CF\end{matrix}\right.\)

=> △DCE=△DCF (c.c.c)

b) △DCE=△DCF => \(\widehat{DCE}=\widehat{DCF}\)

va C ∈ EF

=> \(\widehat{DCE}=\widehat{DCR}=90\)

=> DC⊥EF

c) Xet △EAC va △BFC co:

\(\left\{{}\begin{matrix}\widehat{E}=\widehat{F}\\EC=CF\\\widehat{EAC}=\widehat{CBF}\end{matrix}\right.\)

=> △EAC=△BFC ( g.c.g) => AC=BC

Xet △DAC va △DBC co:

\(\left\{{}\begin{matrix}DA=DB\\\widehat{DAC}=\widehat{DBC}=90\\AC=BC\end{matrix}\right.\)

=> △DAC=△DBC (c.g.c)