Question

ABCD is a parallelogram, X is the midpoint of BC. AX is produced to meet DC produced at Q. ABPQ is a parallelogram.

Prove that

1. ΔABX ≅ ΔQCX
2. DC = CQ = QP

Answer 1

Given:

ABCD is a parallelogram.

X is the midpoint of BC. 

AX is produced to meet DC produced at Q. 

ABPQ is a parallelogram.

To Prove:

1. ΔABX ≅ ΔQCX

2. DC = CQ = QP

Proof:

1. Since ABCD is a parallelogram, AB || DC and AD || BC and opposite sides are equal:

AB = DC and AD = BC.

2. X is the midpoint of BC.

Therefore, BX = XC.

3. Since ABPQ is a parallelogram with vertices A, B, P, Q, AB || PQ and AP || BQ. 

Also, AB = PQ and AP = BQ.

4. AX is produced to meet DC at Q.

So, points A, X and Q are collinear.

5. Consider triangles ABX and QCX:

AB = DC   ...(Opposite sides of parallelogram ABCD)

BX = XC   ...(Since X is midpoint of BC)

∠ABX = ∠QCX   ...(Alternate interior angles, since AB || DC and BX and CX are transversals).

6. By SAS congruence criterion, ΔABX ≅ ΔQCX.

7. From congruency (CPCT - corresponding parts of congruent triangles), AX = QX, ∠AXB = ∠CXQ and so on.

8. Since ABPQ is a parallelogram and AB = PQ, PQ = AB = DC.

9. From the setup and congruent triangles, CQ = DC because Q lies on the extension of DC and since Q is constructed such that ABPQ is a parallelogram, the lengths CQ and QP also equal DC and AB respectively.

So, DC = CQ = QP.