The angle bisector of ∠ACD in rhombus ABCD makes a 64° angle with the diagonal

BD
. Find the measure of ∠BAD.

Assuming you made a sketch.

Let the angle bisector of ∠ACD meet BD at E
let the diagonals intersect at F

In a rhombus the diagonals right-bisect each other.
So, in triangle CEF, angle F = 90
angle E = 64, thus angle FCE = 26°

which makes angle BCD = 104°
and of course ∠BAD = 104°

nice save, Reiny. I later realized how I had misread the problem.

To find the measure of ∠BAD, we need to consider the properties of a rhombus. In a rhombus, the opposite angles are equal.

Let's consider the diagonal BD. Since the angle bisector of ∠ACD makes a 64° angle with the diagonal BD, the opposite angle formed will also be 64°.

So, ∠ACB = 64°.

Now, let's consider the other diagonal, AC. Since ABCD is a rhombus, the diagonals are perpendicular bisectors of each other. This means that ∠ACB is right angled.

Since ∠ACB is a right angle, and ∠ACB = 64°, we can find ∠ABC using the property of supplementary angles. The sum of angles in a triangle is 180°, so ∠ABC + ∠ACB + ∠BAC = 180°.

Substituting the known values, we have:
∠ABC + 64° + ∠BAC = 180°
∠ABC + ∠BAC = 116°

Since ABCD is a rhombus, ∠ABC = ∠BAC. So, we have:
∠ABC = ∠BAC = 116°/2 = 58°

Finally, to find ∠BAD, we know that it is the sum of ∠ABC and ∠BAC. Therefore, ∠BAD = 58° + 58° = 116°.

So, the measure of ∠BAD is 116°.

In summary, to find the measure of ∠BAD, we used the properties of a rhombus and the concept of supplementary angles.

Huh?

The angle bisector is the diagonal AC.

The diagonals of a rhombus are perpendicular.