We are asked to determine whether the following modification of Theorem 6-17 is true or not.
IfABCDis a parallelogram
andACbisects∠ BAD,
thenABCDis a rhombus.
Let's draw the parallelogram and the angle bisector diagonal.
The diagonal divides the angle at A into two congruent angles.
∠ 1≅∠ 2
Recall that the opposite sides of a parallelogram are parallel, so AC is a transversal to two parallel lines.
Using the Transitive Property of Congruence, this means that all sides of the parallelogram are congruent, so it is a rhombus. We are asked to summarize the proof in a paragraph proof.
Completed Proof
2
&Given:&&ABCDis a parallelogram
& &&ACbisects∠ BAD,
&Prove:&&ABCDis a rhombus.
Proof:
If AC bisects ∠ BAD, then ∠ CAD≅∠ BAC. Since opposite sides of parallelogram ABCD are parallel, the alternate interior angles ∠ BAC and ∠ ACD are also congruent. This means that angles ∠ CAD and ∠ ACD are congruent, so triangle △ ACD is isosceles with congruent sides AD and DC. Since opposite sides of a parallelogram are also congruent, this means that all sides of ABCD are congruent, so ABCD is a rhombus.
