What, again?

Since any two tangents to a circle originating from the same point are congruent, we can have $AE=AD= a$ , $BE = BF = b$ , $FC = CD = c$ . Then, we have the following systems of equations:

$a+b = 3$ , $b+c = 4$ , $a+c = 5$

Adding the three together, we get

$2(a+b+c) = 12$ , $a+b+c = 6$

Subtracting the first and second equations respectively from this, we get

$c = 3$ , $a = 2$

Our desired answer is the $ac = \boxed{6}$

Consider a right triangle with incircle radius $r$ . Let the hypotenuse be split into two sections of length $x$ and $y$ as shown. Then we can show that $\begin{cases} { \left( r+x \right) }^{ 2 }+{ \left( r+y \right) }^{ 2 }={ \left( x+y \right) }^{ 2 } \Rightarrow r^2+r(x+y)=xy\\ { \triangle }_{ AREA }=\cfrac { 1 }{ 2 } \left( r+x \right) \left( r+y \right) = \cfrac { 1 }{ 2 } \left( r^2+r(x+y)+xy \right) = \cfrac { 1 }{ 2 } \left( 2xy \right) = xy \end{cases}$ The triangle in question is a right triangle and its area is $\cfrac { 1 }{ 2 } (3)(4) = 6$ . Therefore $|AD| \cdot |DC| = \large \color{#20A900}{\boxed{6}}$ .

Let $O$ be the centre of circle.Then, join $FO$ and $OE$ to get $FOBE$ as square.

And we know that,( See this. )

$\Rightarrow r×s=\text{Area of triangle}$

$r×\dfrac{5+4+3}{2}=\dfrac{1}{2}×b×h$

$r×\dfrac{5+4+3}{2}=\dfrac{1}{2}×4×3$

$r=1$

Now since,

$\Rightarrow EB=BF=EO=FO=1$ .

$AE=AD=3-1=2$ .

Similarly, $FC=DC=4-1=3$ .

Therefore, $AD×DC=2×3=6$ .