Points $D$ and $E$ are chosen on the exterior of $\vartriangle ABC$ such that $\angle  ADC = \angle  BEC = 90^o$. If $\angle ACB = 40^o$, $AD = 7$, $CD = 24$, $CE = 15$, and $BE = 20$, what is the measure of $\angle ABC $ in,degrees?
1. **Identify the right triangles and apply the Pythagorean theorem:**
   - Given that $\angle ADC = 90^\circ$ and $\angle BEC = 90^\circ$, triangles $ADC$ and $BEC$ are right triangles.
   - For $\triangle ADC$, we have $AD = 7$ and $CD = 24$. Using the Pythagorean theorem:
     \[
     AC = \sqrt{AD^2 + CD^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25
     \]
   - For $\triangle BEC$, we have $BE = 20$ and $CE = 15$. Using the Pythagorean theorem:
     \[
     BC = \sqrt{BE^2 + CE^2} = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25
     \]

2. **Determine the properties of $\triangle ABC$:**
   - From the calculations above, we have $AC = 25$ and $BC = 25$. This implies that $\triangle ABC$ is isosceles with $AC = BC$.

3. **Calculate the measure of $\angle ABC$:**
   - Since $\triangle ABC$ is isosceles with $AC = BC$, the base angles $\angle BAC$ and $\angle ABC$ are equal.
   - Given that $\angle ACB = 40^\circ$, the sum of the angles in $\triangle ABC$ is $180^\circ$. Therefore:
     \[
     \angle BAC + \angle ABC + \angle ACB = 180^\circ
     \]
     \[
     \angle BAC + \angle ABC + 40^\circ = 180^\circ
     \]
     \[
     \angle BAC + \angle ABC = 140^\circ
     \]
   - Since $\angle BAC = \angle ABC$, let $\angle BAC = \angle ABC = x$. Then:
     \[
     x + x = 140^\circ
     \]
     \[
     2x = 140^\circ
     \]
     \[
     x = 70^\circ
     \]
   - Therefore, $\angle ABC = 70^\circ$.

The final answer is $\boxed{70^\circ}$.