Question

In the adjoining figure, ABCD is a parallelogram in which P is the mid-point of DC. If Q is any point on AC such that CQ = `1/4` CA and PQ produced meet BC at R, prove that R is the mid-point of BC.

Answer 1

Given:

ABCD is a parallelogram.

P is the midpoint of DC. 

Q lies on AC with CQ = `1/4` × CA. 

The line PQ produced meets BC at R.

To Prove:

R is the midpoint of BC.

Proof [Step-wise]:

1. Put coordinates to the vertices choice of coordinates simplifies algebra:

Let A = (0, 0)

B = (b, 0)

D = (d₁, d₂)

Then, since ABCD is a parallelogram.

C = B + D – A

= (b + d₁, d₂)

2. Find the coordinates of P midpoint of DC:

D = (d₁, d₂), 

C = (b + d₁, d₂)

So `P = ((d_1 + (b + d_1))/2, (d_2 + d_2)/2)`

= `((b + 2d_1)/2, d_2)`

3. Find coordinates of Q given CQ = `1/4` × CA:

Parameterizing AC from A to C.

The point Q satisfying CQ 

= `1/4` × CA is at s   ...`(because 1 - s = 1/4)`

= `3/4` along AC from A

Thus `Q = A + (3/4)(C - A)`

= `(3/4) (b + d_1, d_2)`

= `(3/4 (b + d_1), 3/4 d_2)`

4. Compute the vector Q – P:

`Q - P = (3/4 (b + d_1) - (b + 2d_1)/2, 3/4 d_2 - d_2)`

= `((b - d_1)/4, (-d_2)/4)`

5. Parametric equation of line PQ:

Points on PQ are P + t(Q – P)

= `((b + 2d_1)/2 + (t(b - d_1))/4, d_2(1 - t/4))`

6. Parametric equation of BC:

Points on BC are B + u(C – B) = (b + ud₁, ud₂), 0 ≤ u ≤ 1.

7. Find intersection R by equating coordinates:

From y-coordinates:

`d_2(1 - t/4) = u d_2`

If d₂ ≠ 0 we cancel d₂ to get `u = 1 - t/4`.

From x-coordinates:

`(b + 2d_1)/2 + (t(b - d_1))/4`

= b + ud₁ 

Substitute `u = 1 - t/4`:

`(b + 2d_1)/2 + (t(b - d_1))/4`

= `b + d_1 - (td_1)/4`

Multiply both sides by 4 and simplify:

2(b + 2d₁) + t(b – d₁) 

= 4b + 4d₁ – td₁

⇒ 2b + 4d₁ + tb – td₁ = 4b + 4d₁ – td₁

⇒ 2b + tb = 4b 

⇒ tb = 2b

Assuming b ≠ 0 (B ≠ A), we get t = 2.

Then `u = 1 - t/4`

= `1 - 2/4`

= `1/2`

8. Interpretation:

`u = 1/2` means the intersection R corresponds to the midpoint of BC it divides BC in the ratio 1 : 1. 

Therefore, R is the midpoint of BC.