Given: ABCD is arhombus.

prove:AC bisects angle BAD and angel BCD.

To prove that AC bisects angle BAD and angle BCD in the rhombus ABCD, we need to show that angle BAC is equal to angle CAD and angle ACB is equal to angle BCD.

Proof:

Step 1: Draw the diagonal BD in rhombus ABCD.
Step 2: Since ABCD is a rhombus, we know that all sides are equal and opposite angles are equal.
Step 3: Therefore, we have AB = BC, BC = CD, CD = DA, and DA = AB.
Step 4: In triangle ABD and triangle BCD, we have AB = BC (from step 3), BD is a common side, and AD = CD (opposite sides in the rhombus are equal).
Step 5: By the side-side-side (SSS) congruence postulate, triangle ABD is congruent to triangle BCD.
Step 6: As a result, angle BAC is congruent to angle CAD (corresponding parts of congruent triangles are congruent).
Step 7: Similarly, angle ACB is congruent to angle BCD.
Step 8: Hence, AC bisects angle BAD and angle BCD, as it divides them into two congruent angles.

Therefore, we have proven that AC bisects angle BAD and angle BCD in the rhombus ABCD.