Two parallel lines l and m are cut by a transversal at A and D. In an isosceles triangle ABC, side AB lies along l and side AC lies along the transversal, with AB = AC. The angle at D corresponding to ∠BAC measures 40°. Find ∠ACB.

The correct option is 70°.

Step 1: Determine the measure of ∠BAC using parallel line properties

We are given that two parallel lines l and m are intersected by a transversal line passing through points A and D. When two parallel lines are cut by a transversal, corresponding angles are equal.

Since the angle at D corresponding to ∠BAC measures 40°, the measure of ∠BAC is:

$\angle BAC=40°$

Step 2: Apply the properties of an isosceles triangle

In triangle ABC, side AB lies along line l and side AC lies along the transversal line. We are given that triangle ABC is isosceles with:

$AB=AC$

In an isosceles triangle, the angles opposite the equal sides are equal in measure. Therefore, the base angles opposite to sides AB and AC are equal:

$\angle ABC=\angle ACB$

Step 3: Calculate ∠ACB using the angle sum property

The sum of all interior angles in any triangle is always 180°:

$\angle BAC+\angle ABC+\angle ACB=180°$

Substitute ∠BAC = 40° and replace ∠ABC with ∠ACB:

$40°+\angle ACB+\angle ACB=180°$

$40°+2\angle ACB=180°$

Subtract 40° from both sides of the equation:

$2\angle ACB=180°-40°$

$2\angle ACB=140°$

Divide both sides by 2:

$\angle ACB=\frac{140°}{2}$

$\angle ACB=70°$

Therefore, the measure of ∠ACB is 70°.