Question

In ΔABC, ∠B = 90°. M and N are mid-points of AB and BC respectively.

Prove that

1. CM² + AN² = 5MN²
2. AN² + CM² = AC² + MN²

Answer 1

Given:

In ΔABC, ∠B = 90°.

M and N are mid-points of AB and BC respectively.

To Prove:

1. CM² + AN² = 5MN²
2. AN² + CM² = AC² + MN²

Proof:

Let the right-angled triangle △ABC be such that:

AB = 2y, BC = 2x

Then AC² = (2x)² + (2y)² = 4(x² + y²)

Since M and N are midpoints:

AM = MB = y and BN = NC = x

Step 1: Find coordinates of points:

Let B(0, 0), C(2x,0) and A(0, 2y).

Then M(0, y), N(x, 0)

Step 2: Find the required distances:

CM² = (2x − 0)² + (0 − y)² = 4x² + y²

AN² = (0 − x)² + (2y − 0)² = x² + 4y²

MN² = (x − 0)² + (0 − y)² = x² + y²

Step 3: Prove (i):

CM² + AN² = (4x² + y²) + (x² + 4y²)

CM² + AN² = 5(x² + y²)

But MN² = x² + y²,

CM² + AN² = 5MN²​

Hence, (i) is proved.

Step 4: Prove (ii):

AN² + CM² = 5(x² + y²)

and

AC² = 4(x² + y²), MN² = x² + y²

AC² + MN² = 4(x² + y²) + (x² + y²) = 5(x² + y²)

Thus,

AN² + CM² = AC² + MN²

Hence, (ii) is proved.