Question

ABCD is a rhombus. RABS is a straight line such that RA = AB = BS. Prove that RD and SC when produced meet at right angles.

Answer 1

Given: ABCD is a rhombus. R, A, B, S are collinear in that order and RA = AB = BS.

To Prove: Lines RD and SC when produced are perpendicular.

Proof (Step-wise):

1. Let AB = a.

Place a coordinate system on the line RABS:

Take A = (0, 0) and B = (a, 0).

Then, because RA = AB = BS = a and the points lie in the order R–A–B–S, we have R = (–a, 0) and S = (2a, 0).

2. Let vector AD = (p, q). 

Because ABCD is a rhombus, AD = AB in length.

So, p² + q² = a². 

Then D = A + AD

= (p, q) 

C = B + AD

= (a + p, q)

3. Compute vectors along the lines to be checked: 

RD = D – R

= (p – (–a), q – 0)

= (p + a, q)

SC = C – S

= (a + p – 2a, q – 0)

= (p – a, q)

4. Compute the dot product RD · SC: 

RD · SC 

= (p + a)(p – a) + q·q

= (p² – a²) + q² 

= (p² + q²) – a²

Using p² + q² = a² from step 2, we get

RD · SC

= a² – a²

= 0

5. A zero dot product means RD is perpendicular to SC.

RD and SC when produced meet at right angles.