In an isosceles triangle ABC, the base AC = 4, sinA = 2√2 / 3 Find the line segment connecting

In an isosceles triangle ABC, the base AC = 4, sinA = 2√2 / 3 Find the line segment connecting the midpoints of the sides AC and BC.

Let’s call the segment connecting the midpoints of the AC and BC sides the MK segment.
By definition, it is the middle line of the triangle, which means that it is equal to half of the AB side (and half of the BC side too).
Let us lower the height BE to the base of the speaker. It divides the base into two equal parts, therefore, AE = 2.
Thus, we got a right-angled triangle AEB with right angle E. In it we know two angles A and E and side AE. We find the side AB, using the definition of the cosine (this is the ratio of the adjacent leg to the hypotenuse):
AE: AB = CosA = 1-Sin (A) ^ 2
AB = AE: (1-Sin (A) ^ 2)
AB = 2: (1- {2 * (√2) / 3} ^ 2) = 2 / (1-8 / 9) = 2 * 9 / (9-8) = 18
Segment MK = AB / 2 = 18/2 = 9