Question

In ΔАBC, ∠ABC = 90°. X and Y are mid-points of the sides AB and BC respectively.

Prove that:

1. CX² + AY² = 5XY²
2. 4(CX² + AY²) = 5AC²

Answer 1

Given:

In triangle ABC, ∠ABC = 90°.

X and Y are mid-points of AB and BC respectively.

To Prove:

1. CX² + AY² = 5XY²
2. 4(CX² + AY²) = 5AC²

Proof [Step-wise]:

1. Place the triangle on a coordinate plane with B at the origin.

AB along the x-axis and BC along the y-axis. 

Let AB = a and BC = c.

Then 

B = (0, 0)

A = (a, 0)

C = (0, c)

2. Coordinates of midpoints:

X, midpoint of AB:

`X = (a/2, 0)`

Y, midpoint of BC:

`Y = (0, c/2)`

3. Compute squared distances:

`CX^2 = (a/2 - 0)^2 + (0 - c)^2`

= `a^2/4 + c^2`

`AY^2 = (a - 0)^2 + (0 - c/2)^2`

= `a^2 + c^2/4`

Sum: `CX^2 + AY^2 = (a^2/4 + a^2) + (c^2 + c^2/4)` 

= `(5/4)a^2 + (5/4)c^2`

= `(5/4) (a^2 + c^2)`

4. By the Pythagorean theorem 

Since ∠B = 90° 

a² + c² = AC²

Hence, `CX^2 + AY^2 = (5/4) xx AC^2`.

5. Compute XY²:

`XY^2 = (a/2 - 0)^2 + (0 - c/2)^2`

= `a^2/4 + c^2/4`

= `(a^2 + c^2)/4`

= `(AC^2)/4`

6. Combine results:

From step 4 and step 5,

`CX^2 + AY^2 = (5/4) xx AC^2`

= `5 xx ((AC^2)/4)`

= 5 × XY²

This proves (i).

Multiplying CX² + AY²

= `(5/4) xx AC^2` by 4 gives

4(CX² + AY²) = 5 × AC² 

This proves (ii).