Question

In the adjoining figure, ABCD is a squre and R is the mid-point of AB. PQ is any line segment passing through R which meets AD at P and CB produced at Q. Prove that R is the midpoint of PQ.

Answer 1

Given: ABCD is a square and R is the midpoint of AB.

PQ is any line through R meeting AD at P and the extension of CB at Q.

To Prove: R is the midpoint of PQ.

Proof [Step-wise]:

1. Choose a convenient coordinate system:

Let A(0, 0), B(2, 0), C(2, 2), D(0, 2).

Then AB is horizontal of length 2 and R, the midpoint of AB, has coordinates R(1, 0).

Any side length 2k works; 2 is chosen for simplicity.

2. Let the line PQ through R meet AD (x = 0) at P and the produced CB (x = 2) at Q.

Thus, P has coordinates P(0, p) for some p and Q has coordinates Q(2, q) for some q.

3. Collinearity of P, R, Q they lie on the same line implies the slope from P to R equals the slope from R to Q.

Compute slopes:

Slope (PR)

= `(0 - p)/(1 - 0)` 

= –p

Slope (RQ)

= `(q - 0)/(2 − 1)` 

= q 

Hence, –p = q, so q = –p.

4. The midpoint M of PQ has coordinates M 

= `((0 + 2)/2, (p + q)/2)` 

= `(1, (p + q)/2)`

Using q = –p gives `(p + q)/2 = (p + (-p))/2 = 0`.

So, M = (1, 0).

5. But R = (1, 0).

Therefore, M = R.

So, R is the midpoint of PQ.