Angle between cevians

The required angle will always be same for any traingle ABC that satisfies the given condition.

So take a right angle triangle ABC with length AB just greater than length BC, i.e., (AB-BC) tends to an infinitesimally small number (very close to zero). The angle BAC will be 45 degrees.

Therefore the point D and E will be very close to point B and C respectively.

Hence, the angle between AE and CD will be equal to angle ACB which is $\boxed{45}$ degrees.

Let $CB = a$ and $BA = c$ . Then, $DA = a$ and $DB = c - a$ ; $CE = c -a$ and $EB = 2a - c$ .

By Pythagorean Theorem, ${ EA }^{ 2 }=4{ a }^{ 2 }-4ac+2{ c }^{ 2 }$ and ${ CD }^{ 2 }=2{ a }^{ 2 }-2ac+ { c }^{ 2 }$ , so ${ EA }^{ 2 }=2{ CD }^{ 2 }$ .

Draw a parallel line to $AE$ through $D$ . Let this line intersect $CB$ at $F$ . By similar triangles, we get

$FD ={ EA }\cdot \frac { c-a }{ c } =\sqrt { 2 } CD\cdot \frac { c-a }{ c }$ and

$FB=\frac { (c-a)(2a-c) }{ c }$ .

Thus, $FC = a - FB = a - \frac { (c-a)(2a-c) }{ c } = \frac { 2{ a }^{ 2 }-2ac+ { c }^{ 2 } }{ c } = \frac { { CD }^{ 2 } }{ c }$

Because our parallel line, the asked angle and angle CDF has same measure. Let $\angle CDF = \alpha$ . By cosine rule in $\triangle CDF$ we get:

${ CF }^{ 2 }={ CD }^{ 2 }+{ FD }^{ 2 }-2\cdot CD\cdot FD\cdot cos{ \alpha }$

$\frac { { CD }^{ 4 } }{ { c }^{ 2 } } ={ CD }^{ 2 }+2{ CD }^{ 2 }\cdot \frac { { (c-a) }^{ 2 } }{ { c }^{ 2 } } -2\cdot CD\cdot \sqrt { 2 } \cdot CD\cdot \frac { c-a }{ c } cos{ \alpha }$

dividing both sides by ${CD}^{2}$ and multiplying both sides by ${c}^{2}$ we get:

${ CD }^{ 2 }={ c }^{ 2 }+2{ (c-a) }^{ 2 }-2 \sqrt { 2 } (c-a) c\cdot cos{ \alpha }$

Since ${ CD }^{ 2 }=2{ a }^{ 2 }-2ac+ { c }^{ 2 }$ we have:

$2 \sqrt { 2 } (c-a) c\cdot cos{ \alpha }={ c }^{ 2 }+2{ (c-a) }^{ 2 }-2{ a }^{ 2 }+2ac-{ c }^{ 2 }$

$2 \sqrt { 2 } (c-a) c\cdot cos{ \alpha }=2{ c }^{ 2 }-2ac$

$2\sqrt { 2 } (c-a)c\cdot cos{ \alpha }=2c(c-a)$

Then, $\sqrt { 2 } cos{ \alpha }=1$ and thus, $cos{ \alpha }=\frac { 1 }{ \sqrt { 2 } } =\frac { \sqrt { 2 } }{ 2 }$

therefore, $\alpha = \boxed{45°}$ .