The diagonal AB of the rhombus is equal to 3 times the length of AO.
Let's analyze the given information step by step:
We know that angle BAC = 35° and BCD = 139°.
Since ABCD is intersect a yhombus at O, the angles LABO and CDA are congruent as they are corresponding angles formed by parallel lines cut by a transversal.
Also, since LABO and PAGE are opposite angles, they are congruent as well.
Let's calculate the measure of angle ABO:
Since angle ABO and angle BCD are supplementary, we have angle ABO = 180° - 139° = 41°.
Now, we can find the measure of angle OAB:
Angle OAB = 180° - 35° - 41° = 104°.
Using the Law of Sines, we can find the ratio between the side lengths of triangle OAB:
(AO/sin(OAB)) = (AB/sin(AOB))
Let AO = x
sin(OAB) = sin(104°)
sin(AOB) = sin(180° - 104° - 35°)
Using known ratios and substituting values, we can solve for AB and find:
AB = 3x.
So, the diagonal AB of the rhombus is equal to 3 times the length of AO.