Question

ABC is a triangle, right angled at B. M is a point on BC. Prove that AM² + BC² = AC² + BM².

Answer 1

Given that, ABC is a triangle, right-angled at B. M is a point on BC.

The pictorial form of the given problem is as follows,

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
In a right angled triangle ΔABM, by Pythagoras theorem we get,

AM² = AB² + BM²

⇒ AB² = AM² – BM²   ...(i)

Now, we consider the ΔABC, by Pythagoras theorem we get,

AC² = AB² + BC²

⇒ AB² = AC² – BC²   ...(ii)

From (i) and (ii), we get

AM² – BM² = AC² – BC²

AM² + BC² = AC² + BM²

Hence proved.