ABCD is a parallelogram. P is a point on AD such that AP =  AD and Q is a point on BC such that CQ =   BC. Then AQCP is a

The given figure is a parallelogram ABCD. It is given that AP = AD and CQ = BC.

Step 1: Since AP = AD, triangle APD is an isosceles triangle. Therefore, ∠APD = ∠ADP.

Step 2: Similarly, since CQ = BC, triangle BCQ is an isosceles triangle. Therefore, ∠BCQ = ∠BQC.

Step 3: In a parallelogram, opposite angles are equal. Therefore, ∠ADP = ∠BCQ.

Step 4: From steps 1, 2, and 3, we can conclude that ∠APD = ∠BQC.

Step 5: Also, ∠APQ = ∠CPQ (vertically opposite angles).

Step 6: Therefore, by the Angle-Angle (AA) criterion, triangle APQ is similar to triangle CPQ.

Step 7: In similar triangles, corresponding sides are in proportion. Therefore, AP/CP = PQ/PQ = AQ/CQ.

Step 8: But it is given that AP = AD and CQ = BC. Therefore, AP/CP = AD/BC.

Step 9: In a parallelogram, opposite sides are equal. Therefore, AD/BC = 1.

Step 10: From steps 7, 8, and 9, we can conclude that AQ/CQ = 1. Therefore, AQ = CQ.

Step 11: Therefore, AQCP is a rhombus (since all sides are equal).