The given circuit diagram is shown as :


Applying KVL in loop $A B D A$, we get
$2 I_1+4-I_2=0$
$I_2=2 I_1+4$ ...(i)
Applying KVL in loop $B C D B$, we get
$1\left(I_1+I_3\right)-4\left(I_2-I_3\right)-4=0$
$\begin{array}{lr}\Rightarrow & I_1+I_3-4 I_2+4 I_3-4=0 \\ \Rightarrow & I_1+5 I_3-4\left(2 I_1+4\right)-4=0\end{array}$
$\Rightarrow \quad-7 I_1+5 I_3=20$
$\Rightarrow \quad I_3=\frac{20+7 I_1}{5}$ ...(ii)
By applying $\mathrm{KVL}$ in loop $A D C A$, we get
$I_2+4\left(I_2-I_3\right)-10=0$
$\Rightarrow \quad 5 I_2-4 I_3=10$
$\Rightarrow \quad 5\left(2 I_1+4\right)-4\left(\frac{20+7 I_1}{5}\right)=10$ [From Eqs. (i) and (ii)]
$\Rightarrow \quad I_1=1.364 \mathrm{~A}$
From Eq. (i), $I_2=2 \times 1.364+4=6.278 \mathrm{~A}$
From Eq (ii), $I_3=\frac{20+7 \times 1.364}{5}=5.91 \mathrm{~A}$