Answer:
Follows are the solution to this question:
Step-by-step explanation:
In the given question an attached file is missing, which can be defined as follows please find it.
In part (i):
In the [tex]\triangle ABC[/tex], the side BD bisector of [tex]\angle B[/tex]:
according to bisector theorem:
[tex]\to \frac{6}{x}=\frac{4}{6}\\\\\to 6 \times 6 = 4 \times x\\\\\to \frac{6 \times 6}{4} = x\\\\\to x =\frac{36}{4} \\\\\to x =9 \\\\[/tex]
In point (ii):
In the [tex]\triangle ADB[/tex], the side BD bisector of [tex]\angle A[/tex]:
according to bisector theorem:
[tex]\to \frac{8}{4}=\frac{6-y}{y}\\\\\to \frac{2}{1}=\frac{6-y}{y}\\\\ \to 2 =\frac{6-y}{y}\\\\\to 2y = 6-y\\\\ \to 2y+y 6\\\\ \to 3y =6 \\\\ \to y = \frac{6}{3}\\\\\to y= 2[/tex]
