Based on the image, solve the math problems.

Okay, let's break down each problem step-by-step. Problem 1: Given: $$\frac{AB}{BC} = \frac{4}{3}$$, $$AC = 21$$. Find $$AB$$ and $$BC$$. Since $$AC = AB + BC$$, we can say $$AB = 4x$$ and $$BC = 3x$$ for some value of $$x$$. Therefore, $$4x + 3x = 21$$ $$7x = 21$$ $$x = 3$$ Now we can find $$AB$$ and $$BC$$: $$AB = 4x = 4(3) = 12$$ $$BC = 3x = 3(3) = 9$$ Answer: $$AB = 12$$, $$BC = 9$$ Problem 2: Given: $$CB < AC$$ by 3 cm, $$AB = 15$$ cm. Find $$AC$$ and $$CB$$. Let $$AC = y$$. Then $$CB = y - 3$$. We know that $$AB = AC + CB$$, so $$15 = y + (y - 3)$$ $$15 = 2y - 3$$ $$18 = 2y$$ $$y = 9$$ Now we find $$AC$$ and $$CB$$: $$AC = y = 9$$ $$CB = y - 3 = 9 - 3 = 6$$ Answer: $$AC = 9$$ cm, $$CB = 6$$ cm Problem 3: Given: $$AB = 12$$, $$AM = 8$$. Find $$MB$$. Since $$AB = AM + MB$$, we can find $$MB$$ as follows: $$12 = 8 + MB$$ $$MB = 12 - 8$$ $$MB = 4$$ Answer: $$MB = 4$$ Problem 4: Given: $$M$$ is the midpoint of $$AK$$, $$AB = 20$$. Find $$AK$$. Oops, it appears that some information is missing or the question itself is formulated incorrectly. Need to know the relationship between $$AK$$ and $$AB$$. Perhaps $$AK$$ is somehow related to $$AB$$ or there's a typo. If we assume, that the midpoint of AB is M then: If M is the midpoint of AB, then AM = MB = AB / 2 = 20 / 2 = 10 Since M is the midpoint of AK, then AM = MK. Therefore, AK = 2 * AM = 2 * 10 = 20. So, AK = 20. Answer: Assuming M is midpoint of AB, and also the midpoint of AK, then AK=20