- a) Ta có:
\(\begin{array}{l}AB \bot AC\,\,\left( {gt} \right)\\KE \bot AC\,\,\left( {gt} \right)\\ \Rightarrow AB//KE\end{array}\)
- b) Xét \(\Delta ABC\,\& \Delta KEC\) có:
\(\angle K = \angle A = {90^o}\)
\(\begin{array}{l}EC = CB\left( {gt} \right)\\CK = CA\left( {gt} \right)\end{array}\)
\(\angle ECK = \angle BCA\)(đối đỉnh)
\( \Rightarrow \Delta ABC\, = \Delta KEC\,\,\,\left( {c.g.c} \right)\)
- c) Vì \(\Delta ABC\, = \Delta KEC\,\,\,\left( {c.g.c} \right)\)
\( \Rightarrow AB = KE\) (hai cạnh tương ứng)