Question

If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that : CQ || AP.

Answer 1

Given: ABCD is a parallelogram.

P and Q are points of trisection of the diagonal BD. 

So, B, P, Q, D are collinear and BP = PQ = QD.

To Prove: CQ || AP.

Proof [Step-wise]:

1. Place an origin at B and use position vectors.

Let the vector BD = 3u.

So, B = 0, D = 3u. 

P = B + u = u and Q = B + 2u = 2u.   ...(P and Q trisect BD)

2. Let the position vector of A be a. 

For a parallelogram ABCD,

We have A + C = B + D.   ...(Opposite vertices of a parallelogram satisfy A + C = B + D)

So, C = B + D – A

= 3u – a

3. Compute AP and CQ as vectors:

AP = P – A

= u – a

CQ = Q – C

= 2u – (3u – a)

= a – u

= –(u – a)

4. Since CQ = –AP.   ...(CQ is a scalar multiple –1 of AP)

CQ and AP are collinear and point in opposite directions. 

Hence, CQ || AP.   ...(One nonzero vector is a scalar multiple of another exactly when the corresponding segments are parallel)

Therefore, CQ || AP.