Question

In a square ABCD; A is joined to a point X on BC and D is joined to a point Y on AB. If AX = DY, prove that AX is perpendicular to DY.

Answer 1

Given: ABCD is a square. A is joined to a point X on BC and D is joined to a point Y on AB. AX = DY.

To Prove: AX is perpendicular to DY.

Proof [Step-wise]:

1. Place coordinates:

Let the side of the square be a. 

Put A ≡ (0, 0), B ≡ (a, 0), C ≡ (a, a), D ≡ (0, a).

2. Let X lie on BC.

Then X has coordinates X ≡ (a, t) for some t with 0 ≤ t ≤ a.

3. Let Y lie on AB.

Then Y has coordinates Y ≡ (u, 0) for some u with 0 ≤ u ≤ a.

4. Compute squared lengths:

AX² = (a – 0)² + (t – 0)² 

= a² + t²

DY² = (u – 0)² + (0 – a)²

= u² + a²

5. Given AX = DY.

So, AX² = DY². 

Hence, a² + t² = u² + a², which simplifies to t² = u². 

Because t, u ≥ 0 they measure distances along sides, we get t = u.

6. Therefore, X = (a, t) and Y = (t, 0).

The slope of AX is `(t - 0)/(a - 0) = t/a`.

The slope of DY is `(0 - a)/(t - 0) = -a/t`.

7. Multiply slopes:

`(t/a) xx (-a/t) = -1`

Two lines whose slopes multiply to –1 are perpendicular.

Hence, AX ⟂ DY.