Final answer:
To prove that ABCD is a rhombus, we need to show that all sides are congruent. Using the Angle Bisector Theorem and the fact that diagonal BD divides ABCD into congruent triangles, we can show that AB is congruent to BC and AD is congruent to BC. Therefore, ABCD is a rhombus.
Step-by-step explanation:
To prove that ABCD is a rhombus, we need to show that all sides are congruent. Given that AD is congruent to AB and BC is congruent to CD, we can start by proving that AB is congruent to BC.
Since BD is an angle bisector, we can use the Angle Bisector Theorem which states that in a triangle, an angle bisector divides the opposite side into segments that are proportional to the lengths of the other two sides.
In this case, BD is the angle bisector of ∠B and ∠D, so it divides AC into segments that are proportional to AB and BC. Therefore, AB/BC = AD/CD, which simplifies to AB/BC = 1 since AD = AB and CD = BC. This implies that AB is congruent to BC.
Next, we can use the fact that diagonal BD is also an angle bisector to prove that AD is congruent to BC. Since BD is an angle bisector of ∠B and ∠D, it divides ABCD into two congruent triangles, ABD and CBD. Therefore, AD is congruent to BC.
Finally, since AB is congruent to BC and AD is congruent to BC, we can conclude that all the sides of ABCD are congruent, which means that ABCD is a rhombus.