Angles Parallel Lines F89D07

1. **State the problem:** We have an isosceles triangle $ABC$ with $BA = BC$, and a line $PA$ parallel to $BC$. Given that $\angle ACB = 70^\circ$, we need to find the value of $x$, which is the angle between $PA$ and $AB$ at point $A$. 2. **Recall properties:** In an isosceles triangle, the angles opposite equal sides are equal. Since $BA = BC$, then $\angle BAC = \angle BCA = 70^\circ$. 3. **Identify angles:** We know $\angle ACB = 70^\circ$ (given) and $\angle BAC = 70^\circ$ (from isosceles property). 4. **Calculate $\angle ABC$:** The sum of angles in triangle $ABC$ is $180^\circ$. $$\angle ABC = 180^\circ - \angle BAC - \angle ACB = 180^\circ - 70^\circ - 70^\circ = 40^\circ$$ 5. **Use parallel lines property:** Since $PA \parallel BC$, the alternate interior angles formed by transversal $AB$ are equal. Therefore, $x = \angle ABC = 40^\circ$. 6. **Final answer:** $$x = 40^\circ$$ **Reasons:** - Step 2: Isosceles triangle property. - Step 4: Sum of angles in a triangle is $180^\circ$. - Step 5: Alternate interior angles are equal when lines are parallel. Hence, the value of $x$ is $40^\circ$.