The side BC of ∆ABC is produced to a point D. If AC = BC and ∠BAC = 70°, then find the value of 2.5 ∠ACD − 1.5 ∠ABC.

To find the value of the expression, we can analyze the properties of the given triangle step-by-step.

Step 1: Find the value of $\angle \mathrm{ABC}$
We are given that in $△\mathrm{ABC}$, $\mathrm{AC}=\mathrm{BC}$ and $\angle \mathrm{BAC}=70°$.
Since the two sides $\mathrm{AC}$ and $\mathrm{BC}$ are equal, $△\mathrm{ABC}$ is an isosceles triangle. In an isosceles triangle, the angles opposite to the equal sides are also equal. Therefore:
$\angle \mathrm{ABC}=\angle \mathrm{BAC}$
Given that $\angle \mathrm{BAC}=70°$, we have:
$\angle \mathrm{ABC}=70°$

Step 2: Find the value of the exterior angle $\angle \mathrm{ACD}$
The side $\mathrm{BC}$ of $△\mathrm{ABC}$ is produced to a point $D$, forming the exterior angle $\angle \mathrm{ACD}$ at vertex $C$.
According to the exterior angle theorem, the exterior angle of a triangle is equal to the sum of its two opposite interior angles. Therefore:
$\angle \mathrm{ACD}=\angle \mathrm{BAC}+\angle \mathrm{ABC}$
Substituting the values we know:
$\angle \mathrm{ACD}=70°+70°=140°$

Step 3: Calculate the value of $2.5\angle \mathrm{ACD}-1.5\angle \mathrm{ABC}$
Now we substitute the values of $\angle \mathrm{ACD}=140°$ and $\angle \mathrm{ABC}=70°$ into the given expression:
$2.5\angle \mathrm{ACD}-1.5\angle \mathrm{ABC}=2.5(140°)-1.5(70°)$
Calculating each term:
$2.5\times 140°=350°$
$1.5\times 70°=105°$
Subtracting the values:
$350°-105°=245°$

Thus, the correct option is 245°.