We are given a triangle $ABD$ with a smaller triangle $DEC$ inside. Triangle $DEC$ is equilateral. Angle $BAC$ is given to be $70^\circ$. We are asked to find the size of angle $ABC$.

1. Problem Description

We are given a triangle $ABD$ with a smaller triangle $DEC$ inside. Triangle $DEC$ is equilateral. Angle $BAC$ is given to be $70^\circ$. We are asked to find the size of angle $ABC$.

2. Solution Steps

Since triangle $DEC$ is equilateral, all its angles are equal to $60^\circ$. Thus, $\angle EDC = \angle DCE = \angle DEC = 60^\circ$.

We know that the angles in a triangle add up to $180^\circ$. Therefore, in triangle $ABC$:

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

We know $\angle BAC = 70^\circ$. We can express $\angle ACB$ as the sum of $\angle DCE$ and $\angle ACE$. However, $\angle ACB$ is an exterior angle of triangle $AEC$, so we cannot immediately deduce its value.

Since angles $DCE$ and $ACB$ are supplementary, $\angle DCB = 180^\circ$ if $D$, $C$, and $B$ were collinear. However, we are not given that they are collinear. Instead, we note that $DCB$ form a straight line and so $\angle DCB=180^\circ$. Since $\angle DCE = 60^\circ$, it means $\angle ECB$ is the remaining part of a straight line, i.e., $\angle ECB$ is an exterior angle of $\triangle AEC$.

Then $\angle ACB = \angle ACE$. Since $\angle DCE = 60^\circ$, we have $\angle ACB + \angle DCE = \angle DCB$. However, this does not directly lead to the calculation of $\angle ACB$.

Instead, consider $\angle ADE$ and $\angle ACB$. Since $EDC$ is an equilateral triangle, $ED = DC$.

Consider triangle $ABD$. The sum of angles in triangle $ABD$ is $180^\circ$, that is $\angle DAB + \angle ADB + \angle DBA = 180^\circ$. $\angle BAC$ is the same as $\angle DAB$, so it is equal to $70^\circ$. We are looking for $\angle ABC$ which is the same as $\angle DBA$.

We know $\angle ADB = \angle ADE$. Also, $\angle ACB$ can be written as $\angle ACE + \angle ECB$.

We know that $\angle ACB + \angle ABC + \angle BAC = 180^{\circ}$. Let $\angle ABC = x$. Then $\angle ACB + x + 70^{\circ} = 180^{\circ}$, which means $\angle ACB = 110^{\circ} - x$.

Since the three angles in $\triangle DEC$ are equal to $60^{\circ}$ each, $\angle ACB > \angle DCE$, so we cannot conclude that $\triangle ABC$ is equilateral or isosceles.

We know $\angle ACB$ is the interior angle of $\triangle ABC$.

In $\triangle ABC$, $\angle A + \angle B + \angle C = 180^{\circ}$. We are given $\angle A = 70^{\circ}$, we are looking for $\angle B = x$.

So we have $70^{\circ} + x + \angle C = 180^{\circ}$.

$\angle C = 180^{\circ} - 70^{\circ} - x = 110^{\circ} - x$.

I cannot solve it based on the provided information. However, I am assuming that since DEC is an equilateral triangle, all angles equal 60 degrees. Therefore, $\angle ACB$ is supplementary to $\angle DCE$ if DCE is sitting in the straight line DCB, which seems to be the case. But since the figure is stated as 'Not drawn accurately', this may not be the case either.

However, if D, C, B are on a straight line, then $\angle DCB = 180^{\circ}$. Since $\angle DCE = 60^{\circ}$, then $\angle ACE = 180^{\circ} - 60^{\circ} = 120^{\circ}$.

If we assume $AD = AC$, then $\angle ACB = \angle ADC$. Then, $\angle ABC$ is also equal to $55^{\circ}$.

If $AE=EC$ and $EC=CD$, then $\triangle AED$ is isosceles.

If $AC = BC$, then triangle $ABC$ is isosceles and $\angle BAC = \angle ABC$.

However, the information "The shaded triangle DEC is an equilateral triangle" is useful only if angles $ECA$ and $DCB$ are equal.

Assuming the lines $D, C, B$ are on a straight line. Then $\angle DCB = 180^{\circ}$. Also, we know that $\angle DCE = 60^{\circ}$, so $\angle ECB = 180^{\circ} - 60^{\circ} = 120^{\circ}$. I cannot deduce any more information based on this.

3. Final Answer

I am unable to determine the angle ABC with the given information.