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We see that AC and AB touch the circle at F and D respectively and nowhere else.

Hence, we can call AC and AB the tangents to the circle.

Because tangent is a straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point.

Now, we will use the fact that: Tangents from the same external points are equal in length.

Now tangents AC and AB cut at F and D respectively coming from the same point A.

Hence, \[AF{\text{ }} = {\text{ }}AD\]

Similarly using the same argument again and again, we will have:-

\[BD{\text{ }} = {\text{ }}BE\]

\[CF{\text{ }} = {\text{ }}CE\]

Now, let \[AF = AD = x\] ……..(1)

\[BD = BE = y\] ……..(2)

\[CF = CE = z\;\] ……..(3)

We are already given in the question that AB = 12 cm, BC = 8 cm and AC = 10 cm.

We can clearly see in the figure that \[AB = AD + DB\].

Now, using (1) and (2) in this and the length of AB, we have:-

$ \Rightarrow 12 = x + y$ ……….(4)

We can clearly see in the figure that \[BC = BE + EC\].

Now, using (2) and (3) in this and the length of BC, we have:-

$ \Rightarrow 8 = y + z$ ……….(5)

We can clearly see in the figure that \[AC = AF + FC\].

Now, using (1) and (3) in this and the length of AC, we have:-

$ \Rightarrow 10 = x + z$ ……….(6)

Subtracting (5) from (4), we will have:-

$ \Rightarrow 4 = x - z$ ……..(7)

Adding both (6) and (7), we will get:-

$ \Rightarrow 14 = 2x$

$ \Rightarrow x = 7$ …….(8)

Putting this value of $x$ in (4), we get:-

$ \Rightarrow 12 = 7 + y$

$ \Rightarrow y = 5$ ………(9)

Putting (8) in (6), we get:-

$ \Rightarrow 10 = 8 + z$

$ \Rightarrow z = 2$ ………(9)

Now, we required the lengths of AD, BE and CF.

From (1), (2), (3), (8), (9) and (10), we have:-

\[AD = x = 7cm\]

\[BE = y = 5cm\]

\[CF = z\; = 2cm\]

The students might think that D, E and F are the midpoint but that is not necessary and you cannot put that argument without any solid reason as well.