Question

Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R. Prove that \(AQ = QR\).

Hint:

In parallelograms, extending a side and using a midpoint often creates congruent triangles.

Answer 1

Step 1: Considering the Required Triangles:
We compare triangles ADQ and RCQ to prove AQ = QR.

Step 2: Given and Known Facts:
Q is the midpoint of CD.
Therefore,
DQ = QC

Since ABCD is a parallelogram,
AD ∥ BC

Line R lies on BC, so RC is along BC.

Step 3: Comparing Angles:
∠ADQ = ∠RCQ
(Alternate interior angles, since AD ∥ BC)

Also,
∠AQD = ∠RQC
(Vertically opposite angles)

Step 4: Applying Congruence Rule:
In triangles ADQ and RCQ:
DQ = QC
∠ADQ = ∠RCQ
∠AQD = ∠RQC

Therefore,
ΔADQ ≅ ΔRCQ
(by ASA congruence rule)

Step 5: Using CPCT:
Corresponding sides are equal.

Hence,
AQ = QR
and AD = CR

Final Answer:
By ASA congruence, AQ = QR.