Chứng minh tam giác ABH=tam giác ACK biết ABC cân tại A, BD=CE

\(\Delta ABC\) cân tại A

\(\Leftrightarrow\left\{{}\begin{matrix}AB=AC\\\widehat{ABC}=\widehat{ACB}\end{matrix}\right.\)

Ta có :

\(\left\{{}\begin{matrix}\widehat{DBA}+\widehat{ABC}=180^0\left(kềbuf\right)\\\widehat{ECA}+\widehat{ACB}=180^0\left(kềbuf\right)\end{matrix}\right.\)

Mà \(\widehat{ABC}=\widehat{ACB}\)

\(\Leftrightarrow\widehat{DBA}=\widehat{ECA}\)

Xét \(\Delta ADB;\Delta ACE\) có :

\(\left\{{}\begin{matrix}AB=AC\\\widehat{DBA}=\widehat{ACE}\\DB=CE\end{matrix}\right.\)

\(\Leftrightarrow\Delta ADB=\Delta ACE\left(ch-gn\right)\)

\(\Leftrightarrow\widehat{D}=\widehat{E}\)

Xét \(\Delta HDB;\Delta KEC\) có :

\(\left\{{}\begin{matrix}\widehat{DHB}=\widehat{EKC}=90^0\\DB=CE\\\widehat{D}=\widehat{E}\end{matrix}\right.\)

\(\Leftrightarrow\Delta HDB=\Delta KEC\left(ch-gn\right)\)

\(\Leftrightarrow HB=KC\)

b/ \(\Delta ADB=\Delta ACE\left(cmt\right)\)

\(\Leftrightarrow\widehat{DAB}=\widehat{CAE}\)

Xét \(\Delta ABH;\Delta ACK\) có :

\(\left\{{}\begin{matrix}\widehat{AHB}=\widehat{AKC}=90^0\\\widehat{HAB}=\widehat{KAC}\\AB=AC\end{matrix}\right.\)

\(\Leftrightarrow\Delta ABH=\Delta ACK\left(ch-gn\right)\)