10. It is the incenter, the perpendicular from point $I$ to $DE$ intersects $DE$ at point $M, MI$ extended intersects $BC$ at point $N$, if $IN=2IM$, find $\angle A$.
Let $\triangle ABC$ have three corresponding sides $a, b, c$, and three corresponding angles $\angle A, \angle B, \angle C$. Suppose $\angle IED = \alpha, \angle IDE = \beta$. For $\triangle INC$, by the Law of Sines, we have:
\[
\frac{IN}{\sin \frac{C}{2}} = \frac{IC}{\sin \angle INC} = \frac{IC}{\sin \left(\angle NIC + \frac{\angle C}{2}\right)} = \frac{IC}{\sin \left(90^\circ - \alpha + \frac{C}{2}\right)}
\]
By the angle bisector property, we know:
\[
\frac{EI}{IC} = \frac{AE}{AC} = \frac{BE}{BC} = \frac{AB}{AC + BC} = \frac{c}{a + b}
\]
Also, $IN = 2IM = 2EI \sin \alpha$. Therefore:
\[
\frac{\sin \frac{C}{2}}{\sin \left(90^\circ - \alpha + \frac{C}{2}\right)} = \frac{IN}{IC} = \frac{2EI \sin \alpha}{IC} = 2 \sin \alpha \frac{\sin C}{\sin A + \sin B} = 2 \sin \alpha \frac{\sin \frac{C}{2} \cos \frac{C}{2}}{\sin \frac{A+B}{2} \cos \frac{A-B}{2}}
\]
Since $\sin \frac{A+B}{2} = \cos \frac{C}{2}$, the above equation simplifies to:
\[
2 \sin \alpha \sin \left(90^\circ - \alpha + \frac{C}{2}\right) = \cos \frac{A-B}{2}
\]
Using the product-to-sum identities, we get:
\[
\cos \left(90^\circ - 2\alpha + \frac{C}{2}\right) - \cos \left(90^\circ + \frac{C}{2}\right) = \cos \frac{A-B}{2}
\]
Thus:
\[
\sin \left(2\alpha - \frac{C}{2}\right) = \cos \frac{A-B}{2} - \sin \frac{C}{2} = \cos \frac{A-B}{2} - \cos \frac{A+B}{2} = 2 \sin \frac{A}{2} \sin \frac{B}{2}
\]
Similarly:
\[
\sin \left(2\beta - \frac{B}{2}\right) = 2 \sin \frac{A}{2} \sin \frac{C}{2}
\]
Since $\alpha + \beta = 90^\circ - \frac{1}{2} \angle A = \frac{1}{2} (\angle B + \angle C)$, adding (1) and (2) and using the sum-to-product identities, we get:
\[
\sin \left(\alpha + \beta - \frac{B+C}{4}\right) \cos \left(\alpha - \beta + \frac{B-C}{4}\right) = 2 \sin \frac{A}{2} \sin \frac{B+C}{4} \cos \frac{B-C}{4}
\]
Simplifying, we have:
\[
\cos \left(\alpha - \beta + \frac{B-C}{4}\right) = 2 \sin \frac{A}{2} \cos \frac{B-C}{4}
\]
Subtracting (1) from (2) and using the sum-to-product identities, we get:
\[
\sin \left(\alpha - \beta + \frac{B-C}{4}\right) \cos \left(\alpha + \beta - \frac{B+C}{4}\right) = 2 \sin \frac{A}{2} \sin \frac{B-C}{4} \cos \frac{B+C}{4}
\]
Simplifying, we have:
\[
\sin \left(\alpha - \beta + \frac{B-C}{4}\right) = 2 \sin \frac{A}{2} \sin \frac{B-C}{4}
\]
Squaring (3) and (4) and adding them, we get:
\[
1 = 4 \sin^2 \frac{A}{2}
\]
Thus:
\[
\sin \frac{A}{2} = \frac{1}{2}, \quad \angle A = 60^\circ
\]