A triangle is formed by the lines whose equations are $AB:x+y-5=0$, $BC:x+7y-7=0$ and $CA:7x+y+14=0.$

The slopes of the lines AB, BC and CA are $-1,-\frac{1}{7}\text{ and }-7$ respectively

$\text{ Let }{m}_1=-\frac{1}{7},m_2=-1,m_3=-7$

$\text{ - }{m}_1>m_2>m_3$

tangent of internal angles of the triangle are

$\tan A=\frac{m_1-m_2}{1+m_1{m}_2},\tan B=\frac{m_2-m_3}{1+m_2{m}_3}, \tan C=\frac{m_3-m_1}{1+m_1{m}_3}$

$\tan A=\frac{3}{4},\tan B=\frac{3}{4}\text{ and }\tan C=-\frac{24}{7}$

interior angles A and B are acute and interior angle C is obtuse

internal bisector of B = acute bisector of B

3x + 6y - 16 = 0 External bisector of C

acute bisector of C

8x + 8y + 7 = 0