Construct a triangle with a perimeter of $10.4 \, cm$ and two base angles of $45^{\circ}$ and $120^{\circ}$. Provide the justification for the construction.

(N/A) Steps of construction:
$1.$ Draw a line segment $XY = 10.4 \, cm$.
$2.$ Construct $\angle LXY = 45^{\circ}$ and $\angle MYX = 120^{\circ}$ at points $X$ and $Y$ respectively.
$3.$ Draw the angle bisectors of $\angle LXY$ and $\angle MYX$. Let these bisectors intersect at point $A$.
$4.$ Draw the perpendicular bisector of $AX$,which intersects $XY$ at point $B$.
$5.$ Draw the perpendicular bisector of $AY$,which intersects $XY$ at point $C$.
$6.$ Join $AB$ and $AC$. Thus,$\triangle ABC$ is the required triangle.
Justification:
Since $B$ lies on the perpendicular bisector of $AX$,$XB = AB$. Similarly,since $C$ lies on the perpendicular bisector of $AY$,$YC = AC$.
The perimeter of $\triangle ABC = AB + BC + AC = XB + BC + CY = XY = 10.4 \, cm$.
Also,$\angle XAB = \angle AXB$ (as $XB = AB$) and $\angle YAC = \angle AYC$ (as $YC = AC$).
By the exterior angle property,$\angle ABC = \angle XAB + \angle AXB = 2 \angle XAB = \angle LXY = 45^{\circ}$.
Similarly,$\angle ACB = \angle YAC + \angle AYC = 2 \angle YAC = \angle MYX = 120^{\circ}$.