The shaded triangle $DEC$ is an equilateral triangle. The angle $BAC$ is $50^{\circ}$. We want to find the size of angle $ABC$.

1. Problem Description

The shaded triangle $DEC$ is an equilateral triangle. The angle $BAC$ is $50^{\circ}$. We want to find the size of angle $ABC$.

2. Solution Steps

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

Also, the sum of angles on a straight line is $180^{\circ}$.

The line $DCB$ is a straight line.

Therefore, $\angle ACB + \angle DCE = 180^{\circ}$.

$\angle ACB = 180^{\circ} - \angle DCE$

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

In triangle $ABC$, the sum of the angles is $180^{\circ}$.

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

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

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

$\angle ABC = 180^{\circ} - 170^{\circ}$

$\angle ABC = 10^{\circ}$.

3. Final Answer

The size of angle $ABC$ is $10^{\circ}$.