1) in triangle ABC, AD is the angle bisector of < A. If BD = 5, CD = 6 and AB = 8, find AC.

Angle at C is common to triangles

ACD and to ACB
since angles in both triangles add to 180, angle ADC = twice angle ABD
call angle CAD = T
then BAD = T
call angle ABD = A
then angle ADC = 2A

then
sin T / 6 = sin 2A /x
sin 2T / 11 = sin A /x
sin T / 5 = sin (180-2A)/8 = sin 2A / 5

sin 2T = 2 sin T cos T
sin 2A = 2 sin A cos A

sin T/6 = 2 sin A cos A/x
2 sin T cos T/11 = sin A /x
sin T /5 = 2 sin A cos A /5

sin T = 12 sin A cos A /x
sin T = 11 sin A /(2x cos T)
so
12 cos A = 11/ (2 cos T)
cos A cos T = 11/24
and
sin T = 2 sin A cos A
but
sin T = 12 sin A cos A/x
so
12/ x = 2
x = 6
I bet there are easier ways :)

Thanks

No problem. You are welcome.

To find AC, we can use the angle bisector theorem.

The angle bisector theorem states that in a triangle, the angle bisector of a vertex divides the opposite side into segments that are proportional to the lengths of the adjacent sides.

In triangle ABC, AD is the angle bisector of angle A, which means it divides side BC into segments BD and CD. We are given that BD = 5, CD = 6, and AB = 8.

Let's assume that AC = x. According to the angle bisector theorem, we can set up the following proportion:

BD/CD = AB/AC

Substituting the given values, we get:

5/6 = 8/x

To solve for x, we can cross-multiply:

5x = 8 * 6

5x = 48

Divide both sides by 5:

x = 48/5

Therefore, AC ≈ 9.6.