Proof of proposition I.4 in Euclid's Elements ("Base angles of an isosceles triangle are congruent")

Question

I have difficulties in understanding the 5th proposition of Euclid's elements in the first book:

Proposition I.5. The base angles in an isosceles triangle are congruent. If the sides of an isosceles triangle are extended beyond the base, the angles formed under the base are congruent.

Proof.

Let $\triangle ABC$ be isosceles with $AB \cong AC$. Extending $AB$ and $AC$ past the base $BC$, we can invoke Prop. I.3 to choose $D$ on $AB$ and $E$ on $AC$ so that $AD \cong AE$. Form segments $DC$ and $EB$. We have SAS (I.4) for $\triangle ADC \cong \triangle AEB$, since $\angle CAD$ is common. It follows that $DC \cong EB$, $\angle ACD \cong \angle ABE$, and $\angle ADC \cong \angle AEB$.

Since the whole $AD \cong AE$, and the part $AB \cong AC$, [Common Notion 3] implies $BD \cong CE$. Since $\angle BDC \cong \angle BEC$, and $BC$ is common, we have SAS for $\triangle BDC \cong \triangle BEC$.

From there, $\angle CBD \cong \angle BCE$, which proves the second assertion of the proposition. We also have $\angle BCD \cong \angle CBE$. Since $\angle ABE \cong \angle ACD$, [Common Notion 3] gives us $\angle ABC \cong \angle ACB$, which proves the result. $\square$

I don't understand this part:

Since $\angle BDC \cong \angle BEC$, and $BC$ is common, we have SAS for $\triangle BDC \cong \triangle BEC$.

Proposition I.4 which is SAS, states something different:

Proposition I.4 (SAS). If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also equal.

Proposition I.5 in the text I'm reading is considering only $BC$ and not two equal sides and angle comprised between them.

Answer 1

In triangles $BDC$ and $CEB$ you know that:

1) $CD\cong BE$;

2) $\angle BDC\cong\angle CEB$;

3) $BD\cong CE$;

4) $BC$ is in common.

Hence $BDC\cong CEB$ either by SAS (1, 2, 3) or by SSS (1, 3, 4).

EDIT.

Euclid was forced to use SAS because he proved SSS only in later Proposition (I.8). He added that $BC$ is in common probably to make clear to the reader that he was referring to Prop. I.4 (see Heath's comment quoted in Mauro ALLEGRANZA's comment below).