Question

In the adjoining figure, ◻ABCD is a parallelogram. P is the mid-point of CD and DQ || PB meets CB produced at R.

Prove that:

1. AD = `1/2` CR
2. DR = 2 PB

Answer 1

Given: ABCD is a parallelogram.

P is the midpoint of CD. 

DQ || PB and the line DQ meets the extension of CB at R.

To Prove:

1. AD = `1/2` CR
2. DR = 2 PB

Proof [Step-wise]:

1. Set a coordinate system valid because the figure is affine:

Let A = (0, 0), B = (b, 0) with b ≠ 0, and D = (0, d) with d ≠ 0.

Then C = B + D – A = (b, d).

2. P is the midpoint of CD.

So, `P = ((c_x + d_x)/2, (c_y + d_y)/2)`

= `((b + 0)/2, d)`

= `(b/2, d)`

3. Vector PB = B – P

= `((b - b)/2, 0 - d)` 

= `(b/2, -d)` 

Since DQ || PB, the line through D meeting CB produced i.e. the line x = b has parametric form

`R(t) = D + t xx (b/2, -d)`

= `(0 + t xx b/2, d − t xx d)`

The x-coordinate of R must be b.

So, `t xx (b/2) = b`

⇒ t = 2

Hence, `R = D + 2 xx (b/2, -d)`

= (b, d – 2d)

= (b, –d)

4. Compute CR and AD:

C = (b, d),

R = (b, –d) 

⇒ CR is vertical of length |d – (–d)| = 2d.

AD is vertical from A(0, 0) to D(0, d) of length d.

Therefore, AD = d 

= `(1/2) xx (2d)` 

= `(1/2) xx CR`

This proves (i).

5. Compute lengths DR and PB:

PB vector = `(b/2, -d)`

⇒ `|PB|^2 = (b/2)^2 + d^2`

= `b^2/4 + d^2`

DR vector = R − D

= (b, –2d)

⇒ |DR|² = b² + (2d)²

= b² + 4d²

Observe `|DR|^2 = 4 xx (b^2/4 + d^2)`

= 4 × |PB|² 

So, |DR| = 2 × |PB|.

Hence, DR = 2 PB.

This proves (ii).

Using coordinates equivalently by affine geometry / midpoint-parallel reasoning, we obtain AD = `1/2` CR and DR = 2 PB, as required.