Let $$ABC$$ be a triangle and let $$D$$ be a point on the segment $$BC$$ such that $$AD=BC$$. Suppose $$\angle CAD=x^{\circ}$$, $$\angle ABC=y^{\circ}$$ and $$\angle ACB=z^{\circ}$$ and $$x, y, z$$ are in an arithmetic progression in that order where the first term and the common difference are positive integers. Find the largest possible value of $$\angle ABC$$ in degrees.

Let the three angles be in an arithmetic progression with first term $$x$$ and common difference $$d\;(\,d\gt 0\,)$$:

$$y = x + d,\qquad z = x + 2d$$

A point $$D$$ is chosen on $$BC$$ such that $$AD = BC$$. Use the Law of Sines in the two relevant triangles.

Case 1: Triangle $$ABC$$
Opposite the angles $$A, B, C$$ lie the sides $$BC, AC, AB$$ respectively. Applying the sine rule,

$$\frac{BC}{\sin(180^\circ - y - z)} = \frac{AC}{\sin y}$$ \;⇒\; $$BC = AC \,\frac{\sin(y+z)}{\sin y} \qquad -(1)$$ because $$\sin(180^\circ-\theta)=\sin\theta$$.

Case 2: Triangle $$ACD$$
Here the angles are $$\angle CAD = x,\; \angle ACD = z,\; \angle ADC = 180^\circ - x - z$$. Applying the sine rule again,

$$\frac{AD}{\sin z} = \frac{AC}{\sin(180^\circ - x - z)}$$ \;⇒\; $$AD = AC \,\frac{\sin z}{\sin(x+z)} \qquad -(2)$$.

Given $$AD = BC$$, equate the right-hand sides of (1) and (2):

$$AC \,\frac{\sin z}{\sin(x+z)} = AC \,\frac{\sin(y+z)}{\sin y}$$

$$\Longrightarrow\; \sin(x+z)\,\sin(y+z) = \sin y\,\sin z \qquad -(3)$$

Substitute $$y = x+d,\; z = x+2d$$ in (3).

$$x+z = 2(x+d) = 2y,$$ $$y+z = 2y + d.$$

Equation (3) becomes

$$\sin(2y)\,\sin(2y+d) = \sin y\,\sin(y+d) \qquad -(4)$$

Use the identity $$\sin A\sin B = \tfrac12\bigl[\cos(A-B)-\cos(A+B)\bigr]$$ on both products:

$$\sin(2y)\sin(2y+d)=\tfrac12\bigl[\cos d-\cos(4y+d)\bigr],$$ $$\sin y\sin(y+d)=\tfrac12\bigl[\cos d-\cos(2y+d)\bigr].$$

Subtracting, (4) reduces to

$$\cos(2y+d)-\cos(4y+d)=0.$$

For $$\cos\alpha=\cos\beta$$ we must have $$\alpha=\beta+360k$$ or $$\alpha=-\beta+360k$$, $$k\in\mathbb{Z}$$. The first option gives $$2y=-360k$$, impossible for positive $$y$$. Hence use the second:

$$2y+d = -\,(4y+d)+360k$$ $$\Longrightarrow\; 6y+2d = 360k$$ $$\Longrightarrow\; 3y + d = 180k \qquad -(5)$$

Because all angles are acute and lie in a single triangle, only $$k=1$$ is admissible. Thus

$$d = 180 - 3y \qquad -(6)$$

Feasibility conditions

1. $$d\gt0$$ gives $$3y \lt 180\;\Longrightarrow\; y \lt 60.$$

2. The smallest angle $$x = y-d$$ must be positive:    $$x = y - (180-3y) = 4y - 180 \gt 0 \;\Longrightarrow\; y \gt 45.$$

Combining,

$$45 \lt y \lt 60,$$ and $$y$$ is an integer (because $$x,\,d$$ are required integers). So $$y$$ can take each integer from $$46$$ to $$59$$.

The largest possibility is

$$y = 59^\circ.$$

Check the remaining parameters for $$y=59^\circ$$: $$d = 180 - 3(59) = 3^\circ,$$ $$x = 56^\circ,\; z = 62^\circ.$$ Angles $$56^\circ,59^\circ,62^\circ$$ are indeed in arithmetic progression with common difference $$3^\circ$$, and their sum is $$177^\circ$$, leaving $$\angle DAB = d = 3^\circ$$ so that $$\angle A = 56^\circ + 3^\circ = 59^\circ$$ and the triangle’s angles sum to $$180^\circ$$. All construction requirements are satisfied.

Hence the largest possible value of $$\angle ABC$$ is

59°.