3. Given that the two angle bisectors $B D$ and $C E$ of $\triangle A B C$ intersect at point $I, I D=I E, \angle A B C=70^{\circ}$. Then the degree measure of $\angle A$ is $\qquad$ .
3. $40^{\circ}$ or $60^{\circ}$.

When point $D^{\prime}$ does not coincide with $E^{\prime}$, as shown in Figure 8, on $BC$, take $BE^{\prime}=BE$, $CD^{\prime}=CD$, and connect $ID^{\prime}$, $IE^{\prime}$.
It is easy to prove
$\triangle BIE \cong \triangle BIE^{\prime}$,
$\triangle CID \cong \triangle CID^{\prime}$.
Then $IE^{\prime}=IE=ID=ID^{\prime}$.
Thus, $\angle CD^{\prime}I=\angle BE^{\prime}I$.
Also, $\angle BIE^{\prime}=\angle BIE=\angle CID=\angle CID^{\{\prime}$,

Therefore, $\angle ABC=\angle ACB=70^{\circ}$, $\angle A=40^{\circ}$.
Figure 8
When point $D^{\prime}$ coincides with $E^{\prime}$, as shown in Figure 9.
Similarly, by
$$
\begin{array}{l}
\angle BIE^{\prime}=\angle BIE \\
=\angle CID=\angle CID^{\prime},
\end{array}
$$

and $\angle BID=180^{\circ}$, we know
$$
\angle BIC=120^{\circ} \text {. }
$$

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

Figure 9