SOLUTION: I have a diagram of what appears to be a parallelogram, with vertices ABCD with a single bisector going from A to C. angle BAC is = to Angle ACD(alternate angles) and angle A is

Question 890766: I have a diagram of what appears to be a parallelogram, with vertices ABCD with a single bisector going from A to C. angle BAC is = to Angle ACD(alternate angles) and angle A is = to angle C , These are the only markings on the diagram . No sides are marked as being equal. Does this give me enough information to determine if this quad is a parallelogram?
does the bisector and angle markings tell me that I have one set of parallel lines as well as those lines being congruent?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the following drawing is what i think you have based on what you said.

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if i got it right, then you should have enough to declare that ABCD is a parallelogram.

you know that AB is parallel to CD because angle BAC congruent to angle ACD (given) and they are alternate interior angles to AB and CD, therefore AB is parallel to CD.

angle CAD is congruent to angle BCA because of the following:

angle A = angle C
angle A = angle BAC + angle CAD
angle C = angle BCA + angle ACD
therefore angle BAC + angle CAD = angle BCA + angle ACD
you are given that angle BAC = angle ACD
subtract angle BAC from both sides of the equation and angle BAC disappears from the left side of the equation and angle ACD disappears from the right side of the equation and you are left with:
angle CAD = angle BCA
this means that AD is parallel to BC because angle CAD and angle BCA are alternate interior angles of AD and BC and they are congruent. If the alternate interior angles are congruent, then the lines are parallel.

now you have both pairs of opposite sides are parallel which makes ABCD a parallelogram.

you also have triangle ACD congruent to triangle CAB by ASA, with the congruent side between the congruent angles being AC in both triangle congruent to itself by the reflexive property.

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