Question

In the adjoining figure; AP = `1/2` AB and D is the mid-point of AB, Q is the mid-point of PR and DR || BS.

Prove that:

1. AQ || DR
2. PQ = QR = RS

Answer 1

Given:

AP = `1/2` AB.

D is the midpoint of AB.

Q is the midpoint of PR.

DR || BS.

To Prove:

1. AQ || DR.
2. PQ = QR = RS.

Proof [Step-wise]:

1. Choose a convenient coordinate system:

Place B at (0, 0), BC on the x-axis, and let P = (0, p) with p > 0. 

Let R = (r, 0) be the point on BC.

So, PR is the segment from P to R.

Then Q, the midpoint of PR, has coordinates

`Q = ((0 + r)/2, (p + 0)/2)`

= `(r/2, p/2)`

2. Let A lie on PB so A = (0, a) for some a with 0 < a < p. 

Since D is the midpoint of AB, we have `D = (0, a/2)`.

The condition AP = `1/2` AB gives:

AP = p – a,

AB = a

And p – a = `1/2` a 

⇒ p = `3/2` a 

⇒ a = `2/3` p

Hence, `D = (0, a/2)`

= `(0, p/3)`

3. Compute slope of DR:

`D = (0, p/3)`

R = (r, 0)

So, slope (DR) = `(0 - p/3)/(r - 0)` 

= `-p/(3r)`

4. Compute slope of AQ:

A = (0, a) 

= `(0, (2p)/3)` 

`Q = (r/2, p/2)`

Slope (AQ) = `(p/2 - (2p)/3)/(r/2 - 0)`

= `((3p - 4p)/6)/(r/2)`

= `(-p/6) xx (2/r)`

= `-p/(3r)`

Therefore, slope (AQ) = slope (DR). 

So, AQ || DR.

This proves (i).

5. Show PQ = QR = RS:

Q is the midpoint of PR by hypothesis.

So, PQ = QR.

Compute the coordinates of S required by the condition DR || BS.

Lines through B with slope equal to slope (DR) have equation y = m x with `m = -p/(3r)`.

Let S be a point on that line.

If we choose S so that R is the midpoint of QS.

Then S = 2R – Q

= `(2r - r/2, 0 - p/2)`

= `((3r)/2, -p/2)`

The slope of BS is `(-p/2 - 0)/((3r)/2 - 0)`

= `(-p/2)/((3r)/2)`

= `-p/(3r)`

= Slope (DR), so indeed BS || DR and R is the midpoint of QS. 

Hence, QR = RS.

Combining QR = RS with PQ = QR gives PQ = QR = RS.

This proves (ii).