We are given a triangle $ABC$ and a smaller triangle $DEC$ within it. We know that triangle $DEC$ is an equilateral triangle, and that angle $A$ in triangle $ABC$ is $80^{\circ}$. We want to find the measure of angle $ABC$.

1. Problem Description

We are given a triangle $ABC$ and a smaller triangle $DEC$ within it. We know that triangle $DEC$ is an equilateral triangle, and that angle $A$ in triangle $ABC$ is $80^{\circ}$. We want to find the measure of angle $ABC$.

2. Solution Steps

Since $DEC$ is an equilateral triangle, all its angles are $60^{\circ}$. Thus, $\angle DCE = 60^{\circ}$.

Since $AC$ is a straight line segment, the sum of the angles $\angle DCE$ and $\angle ECA$ is $180^{\circ}$.

$\angle DCE + \angle ECA = 180^{\circ}$

$60^{\circ} + \angle ECA = 180^{\circ}$

$\angle ECA = 180^{\circ} - 60^{\circ} = 120^{\circ}$.

Now we know that in triangle $AEC$, the sum of the angles is $180^{\circ}$.

Also, $\angle AEC = 60^{\circ}$, because triangle $DEC$ is equilateral. Then $\angle DEA = 180^{\circ} - 60^{\circ} = 120^{\circ}$.

Consider triangle $ABC$. We know that the sum of its angles is $180^{\circ}$.

$\angle BAC + \angle ACB + \angle ABC = 180^{\circ}$.

We are given $\angle BAC = 80^{\circ}$. We need to find $\angle ACB$.

We know that $\angle ACB = 180^{\circ} - \angle ECA - \angle ECB = 180^{\circ} - (\angle ECA)$.

However, we cannot find the measure of angle $\angle ACB$.

Let's look at $\angle ACE$. $\angle ACE$ is supplementary to $\angle DCE$.

Since $DEC$ is equilateral, $\angle DCE = 60^{\circ}$. Therefore, $\angle ACE = 180^{\circ} - 60^{\circ} = 120^{\circ}$.

Now we can use the angles in triangle $ABC$ equation.

The sum of the angles in triangle $ABC$ is $180^{\circ}$.

$80^{\circ} + \angle ACB + \angle ABC = 180^{\circ}$

$\angle ECA = 180^{\circ} - \angle ACB$. $\angle ACE = 120^{\circ}$.

Therefore, $\angle ECA$ and $\angle ACB$ do not appear to be connected.

Since $DEC$ is an equilateral triangle, $\angle DEC = \angle EDC = \angle ECD = 60^{\circ}$.

We have $\angle BAC = 80^{\circ}$. The sum of angles in triangle ABC is

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0. $\angle ABC + \angle BAC + \angle ACB = 180^{\circ}$

$\angle ACB$ is supplementary to $\angle DCE$, where $\angle DCE = 60^{\circ}$. So it looks like $\angle ACB = 180 - 60=120$.

This is wrong.

It seems there is information missing, as there is no way to relate angle $ACB$ to angle $DEC$.

Let's reconsider. $DEC$ is equilateral, so $\angle DCE=60^{\circ}$. $AC$ and $DC$ form a straight line. So the outside angle $= 180-60 = 120$.

We need to make an assumption, which is that the lines $AE$ and $DB$ are parallel. This would mean that $\angle AED = \angle ABC$. However, this assumption cannot be made.

Let $x$ be the measure of $\angle ABC$. Let $y$ be the measure of $\angle ACB$. Then $80 + x + y = 180$.

$x + y = 100$. $y = 100 - x$. $\angle ACB = y$.

We also know that $\angle DCE = 60^{\circ}$.

Without further assumptions or information, there is not a unique answer.

3. Final Answer

Without more information, we cannot find the size of angle ABC.

There is missing information in the problem statement.