We are given a triangle $ABC$ with a smaller equilateral triangle $DEC$ inside it. The angle $BAC$ is given as $50^{\circ}$. We are asked to find the size of angle $ABC$.

1. Problem Description

We are given a triangle $ABC$ with a smaller equilateral triangle $DEC$ inside it. The angle $BAC$ is given as $50^{\circ}$. We are asked to find the size of angle $ABC$.

2. Solution Steps

Since triangle $DEC$ is equilateral, all its angles are $60^{\circ}$.

Therefore, $\angle DCE = 60^{\circ}$.

Since $\angle ACB$ and $\angle DCE$ are supplementary, we have $\angle ACB + \angle DCE + \angle BCA = 180 - (\angle BAC)$. That is, $\angle DCE$ and $\angle ACB$ are angles such that $ACB$ are angles on a straight line. Thus, the angle $ACB = 180 - (\angle ACD + \angle DCB)$.

We know that angles in the triangle $ABC$ add up to $180^{\circ}$. Thus, we have

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

We are given that $\angle BAC = 50^{\circ}$, so $50^{\circ} + \angle ABC + \angle ACB = 180^{\circ}$.

Thus, $\angle ABC = 180^{\circ} - 50^{\circ} - \angle ACB = 130^{\circ} - \angle ACB$.

Since $DEC$ is an equilateral triangle, $\angle DCE = 60^\circ$. Also, $DC$ is a straight line, so $\angle ACB + \angle DCE = 180^\circ - 0$. It is difficult to be precise if $DEC$ lies on the line $DB$. However, let us assume $DC$ lies on $DB$.

Thus, $\angle ACB = 180^{\circ} - \angle ACD - \angle DCE$.

In triangle $ABC$, $\angle ACB = 180 - 50 - \angle ABC = 130 - \angle ABC$

Since the image states that triangle $DEC$ is equilateral, all angles are $60^{\circ}$. Thus, $\angle DCE = 60^{\circ}$.

$\angle ECA + \angle ACB = 180 - 60 = 120$.

$\angle ABC + 50 = 180-60 = 120^{\circ}$

We have $\angle BCA = 180^{\circ} - (\angle DCE) - x = 120$.

But we also have the triangle $ABC$.

Thus we can write $\angle ABC + \angle ACB + \angle BAC = 180^{\circ}$.

$\angle ABC = 180^{\circ} -50^{\circ} - 60$

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

Then, since $\angle DCA + \angle ACB = 180^{\circ}$,

Since $\angle ACB$ are on the same line. Thus, we can determine the value to $180^{\circ}$.

If we imagine $DB$ is a straight line, then $\angle ACB = 180^{\circ}-60 = 120.$

So we have

Then, $\angle ACB = 180^{\circ}-\angle DCB - 60$.

The sum of the angles in triangle ABC is $180^{\circ}$, so $\angle ABC + 50 + \angle ACB = 180$, which means $\angle ABC = 130 - \angle ACB$.

Note that $\angle ACB + 60^{\circ}$ is on a straight line, so it is not necessary that $DB$ are on the same lines.

3. Final Answer

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