`a)` `BN \cap CE = {I}`

`=> {(9x - 3y - 4 = 0),(x + y - 2 = 0):} <=> {(x = \frac{5}{6}),(y = \frac{7}{6}):} => I (\frac{5}{6};\frac{7}{6})`

`AH` `{(\text{đi qua }I(\frac{5}{6};\frac{7}{6})),(\text{vtpt }\vec{n}(- 5;7)):}`

`=> - 5x + 7y - 4 = 0`

`->` Câu `a)` sai

`b)` `BN \cap BC = {B}`

`=> {(9x - 3y - 4 = 0),(7x + 5y - 8 = 0):} <=> {(x = \frac{2}{3}),(y = \frac{2}{3}):} => B (\frac{2}{3};\frac{2}{3})`

`CE \cap BC = {C}`

`=> {(x + y - 2 = 0),(7x + 5y - 8 = 0):} <=> {(x = - 1),(y = 3):} => C (- 1: 3)`

`=> \vec{BC} = (- \frac{5}{3}; \frac{7}{3})`

`BC \cap AH = {H}`

`=> {(7x + 5y - 8 = 0),(- 5x + 7y - 4 = 0):} <=> {(x = \frac{18}{37}),(y = \frac{34}{37}):} => H (\frac{18}{37};\frac{34}{37})`

Gọi `M (a;b)` là trung điểm của `BC => {(\vec{BM} = (a - \frac{2}{3};b - \frac{2}{3})),(a + 1;b - 3):}`

Vì `M` là trung điểm của `BC` nên 

`=> \vec{BM} + \vec{CM} = 0`

`<=> (a - \frac{2}{3};b - \frac{2}{3}) + (a + 1; b - 3) = 0`

`<=> {(a - \frac{2}{3} + a + 1 = 0),(b - \frac{2}{3} + b - 3 = 0):} <=> {(a = - \frac{1}{66}),(b = \frac{11}{6}):} => M (- \frac{1}{6};\frac{11}{6})`

Gọi `A (c;\frac{5c + 4}{7}) \in AH => \vec{AH} = (\frac{18}{37} - c;\frac{34}{37} - \frac{5c + 4}{7})`

Vì `AH \bot BC` nên `\vec{AH}.\vec{BC} = 0`

`<=> - \frac{5}{3}(\frac{18}{37} - c) + \frac{7}{3}(\frac{34}{37} - \frac{5c + 4}{7}) = 0`

`<=> c = - \frac{4}{5} => A (- \frac{4}{5};0)`

`AM` `{(\text{đi qua }A(- \frac{4}{5};0)),(\text{vtpt} \vec{n} (- \frac{11}{6};\frac{19}{30})):}`

`=> - \frac{11}{6}(x + \frac{4}{5}) + \frac{19}{30}y = 0`

`<=> - \frac{11}{6}x + \frac{19}{30}y - \frac{22}{15} = 0`

`<=> x - \frac{19}{55} + \frac{4}{5} = 0`

`->` Câu `b` sai