- a) Xét \(\Delta ABH = \Delta ACH\) có:
\(\begin{array}{l}AB = AC\left( {gt} \right)\\AH\,\,chung\\BH = CH\left( {gt} \right)\end{array}\)
\( \Rightarrow \Delta ABH = \Delta ACH\,\,\left( {c.c.c} \right)\)
- b) Vì \(\Delta ABH = \Delta ACH\,\,\left( {cmt} \right)\)
\( \Rightarrow \angle BAH = \angle CAH\) (hai góc tương ứng) (1)
Xét \(\Delta ABH\,\,\& \Delta ECH\) ta có:
\(\begin{array}{l}AH = HE\left( {gt} \right)\\HB = HC\left( {gt} \right)\\\angle AHB = \angle EHC\,\,\left( {doi\,dinh} \right)\end{array}\)
\( \Rightarrow \Delta ABH = \Delta EHC\left( {c.g.c} \right)\)
\( \Rightarrow \angle BAH = \angle CEH\) (hai góc tương ứng)
Mà hai góc này ở vị trí so le trong
Do đó: \(AB//EC\)
- c)
Vì
\(\begin{array}{l}\Delta ABH = \Delta CAH\\\Delta ABH = \Delta ECH\\ \Rightarrow \Delta CAH = \Delta CEH\end{array}\)
\( \Rightarrow \angle CAH = \angle CEH\) (hai góc tương ứng)
